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修复mico-sdk错误

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This commit is contained in:
nhkefus
2025-03-11 15:54:45 +08:00
parent 3422912129
commit 2ccb892a1c
2152 changed files with 664341 additions and 702636 deletions
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884
mico-os/platform/RVMDK/Retarget.c
View File

@@ -1,442 +1,442 @@
/***
* File: retarget.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#if defined(__CC_ARM)
#include <string.h>
#include <stdio.h>
#include "mico_platform.h"
#include "platform.h"
#include "common.h"
#include "mico_rtos.h"
#include <rt_misc.h>
#pragma import(__use_no_semihosting_swi)
struct mxchip_mallinfo {
int num_of_chunks; /* number of free chunks */
int total_memory; /* maximum total allocated space */
int allocted_memory; /* total allocated space */
int free_memory; /* total free space */
};
struct __FILE { int handle; /* Add whatever you need here */ };
FILE __stdout;
FILE __stdin;
FILE __stderr;
int fgetc(FILE *f) {
return 0x30;
}
int ferror(FILE *f) {
/* Your implementation of ferror */
return EOF;
}
void _ttywrch(int ch) {
return;
}
void _sys_exit(int return_code) {
label: goto label; /* endless loop */
}
int fputc(int ch, FILE *f) {
MicoUartSend( STDIO_UART, &ch, 1 );
return ch;
}
char *_sys_command_string(char * cmd, int len)
{
cmd = cmd;
len = len;
return 0;
}
// logging.o
char* __iar_Strstr(char* s1, char* s2)
{
int n;
if (*s2)
{
while (*s1)
{
for (n=0; *(s1 + n) == *(s2 + n); n++)
{
if (!*(s2 + n + 1))
return (char *)s1;
}
s1++;
}
return NULL;
}
else
return (char *)s1;
}
//char *__iar_Strrchr(const char *str, int c)
//{
// return strrchr(str, c);
//}
# define STRRCHR __iar_Strrchr
# define STRCHR __iar_Strchr
# define MEMCHR __iar_Memchr
char *
STRRCHR (const char *s, int c)
{
const char *found, *p;
c = (unsigned char) c;
/* Since strchr is fast, we use it rather than the obvious loop. */
if (c == '\0')
return strchr (s, '\0');
found = NULL;
while ((p = strchr (s, c)) != NULL)
{
found = p;
s = p + 1;
}
return (char *) found;
}
char *
STRCHR (const char *s, int c_in)
{
const unsigned char *char_ptr;
const unsigned long int *longword_ptr;
unsigned long int longword, magic_bits, charmask;
unsigned char c;
c = (unsigned char) c_in;
/* Handle the first few characters by reading one character at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
else if (*char_ptr == '\0')
return NULL;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to 8-byte longwords. */
longword_ptr = (unsigned long int *) char_ptr;
/* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
the "holes." Note that there is a hole just to the left of
each byte, with an extra at the end:
bits: 01111110 11111110 11111110 11111111
bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
The 1-bits make sure that carries propagate to the next 0-bit.
The 0-bits provide holes for carries to fall into. */
switch (sizeof (longword))
{
case 4: magic_bits = 0x7efefeffL; break;
default:
abort ();
}
/* Set up a longword, each of whose bytes is C. */
charmask = c | (c << 8);
charmask |= charmask << 16;
if (sizeof (longword) > 4)
/* Do the shift in two steps to avoid a warning if long has 32 bits. */
charmask |= (charmask << 16) << 16;
if (sizeof (longword) > 8)
abort ();
/* Instead of the traditional loop which tests each character,
we will test a longword at a time. The tricky part is testing
if *any of the four* bytes in the longword in question are zero. */
for (;;)
{
/* We tentatively exit the loop if adding MAGIC_BITS to
LONGWORD fails to change any of the hole bits of LONGWORD.
1) Is this safe? Will it catch all the zero bytes?
Suppose there is a byte with all zeros. Any carry bits
propagating from its left will fall into the hole at its
least significant bit and stop. Since there will be no
carry from its most significant bit, the LSB of the
byte to the left will be unchanged, and the zero will be
detected.
2) Is this worthwhile? Will it ignore everything except
zero bytes? Suppose every byte of LONGWORD has a bit set
somewhere. There will be a carry into bit 8. If bit 8
is set, this will carry into bit 16. If bit 8 is clear,
one of bits 9-15 must be set, so there will be a carry
into bit 16. Similarly, there will be a carry into bit
24. If one of bits 24-30 is set, there will be a carry
into bit 31, so all of the hole bits will be changed.
The one misfire occurs when bits 24-30 are clear and bit
31 is set; in this case, the hole at bit 31 is not
changed. If we had access to the processor carry flag,
we could close this loophole by putting the fourth hole
at bit 32!
So it ignores everything except 128's, when they're aligned
properly.
3) But wait! Aren't we looking for C as well as zero?
Good point. So what we do is XOR LONGWORD with a longword,
each of whose bytes is C. This turns each byte that is C
into a zero. */
longword = *longword_ptr++;
/* Add MAGIC_BITS to LONGWORD. */
if ((((longword + magic_bits)
/* Set those bits that were unchanged by the addition. */
^ ~longword)
/* Look at only the hole bits. If any of the hole bits
are unchanged, most likely one of the bytes was a
zero. */
& ~magic_bits) != 0 ||
/* That caught zeroes. Now test for C. */
((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
& ~magic_bits) != 0)
{
/* Which of the bytes was C or zero?
If none of them were, it was a misfire; continue the search. */
const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
if (*cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (sizeof (longword) > 4)
{
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
}
}
}
}
/* Search no more than N bytes of S for C. */
void *
MEMCHR (void const *s, int c_in, size_t n)
{
/* On 32-bit hardware, choosing longword to be a 32-bit unsigned
long instead of a 64-bit uintmax_t tends to give better
performance. On 64-bit hardware, unsigned long is generally 64
bits already. Change this typedef to experiment with
performance. */
typedef unsigned long int longword;
const unsigned char *char_ptr;
const longword *longword_ptr;
longword repeated_one;
longword repeated_c;
unsigned char c;
c = (unsigned char) c_in;
/* Handle the first few bytes by reading one byte at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
--n, ++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
longword_ptr = (const longword *) char_ptr;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
/* Compute auxiliary longword values:
repeated_one is a value which has a 1 in every byte.
repeated_c has c in every byte. */
repeated_one = 0x01010101;
repeated_c = c | (c << 8);
repeated_c |= repeated_c << 16;
if (0xffffffffU < (longword) -1)
{
repeated_one |= repeated_one << 31 << 1;
repeated_c |= repeated_c << 31 << 1;
if (8 < sizeof (longword))
{
size_t i;
for (i = 64; i < sizeof (longword) * 8; i *= 2)
{
repeated_one |= repeated_one << i;
repeated_c |= repeated_c << i;
}
}
}
/* Instead of the traditional loop which tests each byte, we will test a
longword at a time. The tricky part is testing if *any of the four*
bytes in the longword in question are equal to c. We first use an xor
with repeated_c. This reduces the task to testing whether *any of the
four* bytes in longword1 is zero.
We compute tmp =
((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
That is, we perform the following operations:
1. Subtract repeated_one.
2. & ~longword1.
3. & a mask consisting of 0x80 in every byte.
Consider what happens in each byte:
- If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
and step 3 transforms it into 0x80. A carry can also be propagated
to more significant bytes.
- If a byte of longword1 is nonzero, let its lowest 1 bit be at
position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
the byte ends in a single bit of value 0 and k bits of value 1.
After step 2, the result is just k bits of value 1: 2^k - 1. After
step 3, the result is 0. And no carry is produced.
So, if longword1 has only non-zero bytes, tmp is zero.
Whereas if longword1 has a zero byte, call j the position of the least
significant zero byte. Then the result has a zero at positions 0, ...,
j-1 and a 0x80 at position j. We cannot predict the result at the more
significant bytes (positions j+1..3), but it does not matter since we
already have a non-zero bit at position 8*j+7.
So, the test whether any byte in longword1 is zero is equivalent to
testing whether tmp is nonzero. */
while (n >= sizeof (longword))
{
longword longword1 = *longword_ptr ^ repeated_c;
if ((((longword1 - repeated_one) & ~longword1)
& (repeated_one << 7)) != 0)
break;
longword_ptr++;
n -= sizeof (longword);
}
char_ptr = (const unsigned char *) longword_ptr;
/* At this point, we know that either n < sizeof (longword), or one of the
sizeof (longword) bytes starting at char_ptr is == c. On little-endian
machines, we could determine the first such byte without any further
memory accesses, just by looking at the tmp result from the last loop
iteration. But this does not work on big-endian machines. Choose code
that works in both cases. */
for (; n > 0; --n, ++char_ptr)
{
if (*char_ptr == c)
return (void *) char_ptr;
}
return NULL;
}
//mxchipWNET.o
int is_nfc_up(void){
return 0;
}
__weak void *__iar_dlmallinfo(void)
{
return NULL;
}
__weak int aes_decrypt(void)
{
return 0;
}
int nfc_config_stop(void)
{
return 0;
}
int uart_init(void)
{
return 0;
}
int __data_GetMemChunk(void)
{
return 0;
}
char *strdup(const char *src)
{
int len;
char *dst;
if (src == NULL)
return NULL;
if (src[0] == 0)
return NULL;
len = strlen(src) + 1;
dst = (char*)malloc(len);
if (dst)
memcpy(dst, src, len);
return dst;
}
int mico_platform_get_rtc_time (){
return 1;
}
int mico_platform_set_rtc_time (){
return 0;
}
#endif
/***
* File: retarget.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#if defined(__CC_ARM)
#include <string.h>
#include <stdio.h>
#include "mico_platform.h"
#include "platform.h"
#include "common.h"
#include "mico_rtos.h"
#include <rt_misc.h>
#pragma import(__use_no_semihosting_swi)
struct mxchip_mallinfo {
int num_of_chunks; /* number of free chunks */
int total_memory; /* maximum total allocated space */
int allocted_memory; /* total allocated space */
int free_memory; /* total free space */
};
struct __FILE { int handle; /* Add whatever you need here */ };
FILE __stdout;
FILE __stdin;
FILE __stderr;
int fgetc(FILE *f) {
return 0x30;
}
int ferror(FILE *f) {
/* Your implementation of ferror */
return EOF;
}
void _ttywrch(int ch) {
return;
}
void _sys_exit(int return_code) {
label: goto label; /* endless loop */
}
int fputc(int ch, FILE *f) {
MicoUartSend( STDIO_UART, &ch, 1 );
return ch;
}
char *_sys_command_string(char * cmd, int len)
{
cmd = cmd;
len = len;
return 0;
}
// logging.o
char* __iar_Strstr(char* s1, char* s2)
{
int n;
if (*s2)
{
while (*s1)
{
for (n=0; *(s1 + n) == *(s2 + n); n++)
{
if (!*(s2 + n + 1))
return (char *)s1;
}
s1++;
}
return NULL;
}
else
return (char *)s1;
}
//char *__iar_Strrchr(const char *str, int c)
//{
// return strrchr(str, c);
//}
# define STRRCHR __iar_Strrchr
# define STRCHR __iar_Strchr
# define MEMCHR __iar_Memchr
char *
STRRCHR (const char *s, int c)
{
const char *found, *p;
c = (unsigned char) c;
/* Since strchr is fast, we use it rather than the obvious loop. */
if (c == '\0')
return strchr (s, '\0');
found = NULL;
while ((p = strchr (s, c)) != NULL)
{
found = p;
s = p + 1;
}
return (char *) found;
}
char *
STRCHR (const char *s, int c_in)
{
const unsigned char *char_ptr;
const unsigned long int *longword_ptr;
unsigned long int longword, magic_bits, charmask;
unsigned char c;
c = (unsigned char) c_in;
/* Handle the first few characters by reading one character at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0;
++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
else if (*char_ptr == '\0')
return NULL;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to 8-byte longwords. */
longword_ptr = (unsigned long int *) char_ptr;
/* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
the "holes." Note that there is a hole just to the left of
each byte, with an extra at the end:
bits: 01111110 11111110 11111110 11111111
bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
The 1-bits make sure that carries propagate to the next 0-bit.
The 0-bits provide holes for carries to fall into. */
switch (sizeof (longword))
{
case 4: magic_bits = 0x7efefeffL; break;
default:
abort ();
}
/* Set up a longword, each of whose bytes is C. */
charmask = c | (c << 8);
charmask |= charmask << 16;
if (sizeof (longword) > 4)
/* Do the shift in two steps to avoid a warning if long has 32 bits. */
charmask |= (charmask << 16) << 16;
if (sizeof (longword) > 8)
abort ();
/* Instead of the traditional loop which tests each character,
we will test a longword at a time. The tricky part is testing
if *any of the four* bytes in the longword in question are zero. */
for (;;)
{
/* We tentatively exit the loop if adding MAGIC_BITS to
LONGWORD fails to change any of the hole bits of LONGWORD.
1) Is this safe? Will it catch all the zero bytes?
Suppose there is a byte with all zeros. Any carry bits
propagating from its left will fall into the hole at its
least significant bit and stop. Since there will be no
carry from its most significant bit, the LSB of the
byte to the left will be unchanged, and the zero will be
detected.
2) Is this worthwhile? Will it ignore everything except
zero bytes? Suppose every byte of LONGWORD has a bit set
somewhere. There will be a carry into bit 8. If bit 8
is set, this will carry into bit 16. If bit 8 is clear,
one of bits 9-15 must be set, so there will be a carry
into bit 16. Similarly, there will be a carry into bit
24. If one of bits 24-30 is set, there will be a carry
into bit 31, so all of the hole bits will be changed.
The one misfire occurs when bits 24-30 are clear and bit
31 is set; in this case, the hole at bit 31 is not
changed. If we had access to the processor carry flag,
we could close this loophole by putting the fourth hole
at bit 32!
So it ignores everything except 128's, when they're aligned
properly.
3) But wait! Aren't we looking for C as well as zero?
Good point. So what we do is XOR LONGWORD with a longword,
each of whose bytes is C. This turns each byte that is C
into a zero. */
longword = *longword_ptr++;
/* Add MAGIC_BITS to LONGWORD. */
if ((((longword + magic_bits)
/* Set those bits that were unchanged by the addition. */
^ ~longword)
/* Look at only the hole bits. If any of the hole bits
are unchanged, most likely one of the bytes was a
zero. */
& ~magic_bits) != 0 ||
/* That caught zeroes. Now test for C. */
((((longword ^ charmask) + magic_bits) ^ ~(longword ^ charmask))
& ~magic_bits) != 0)
{
/* Which of the bytes was C or zero?
If none of them were, it was a misfire; continue the search. */
const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
if (*cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (sizeof (longword) > 4)
{
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
if (*++cp == c)
return (char *) cp;
else if (*cp == '\0')
return NULL;
}
}
}
}
/* Search no more than N bytes of S for C. */
void *
MEMCHR (void const *s, int c_in, size_t n)
{
/* On 32-bit hardware, choosing longword to be a 32-bit unsigned
long instead of a 64-bit uintmax_t tends to give better
performance. On 64-bit hardware, unsigned long is generally 64
bits already. Change this typedef to experiment with
performance. */
typedef unsigned long int longword;
const unsigned char *char_ptr;
const longword *longword_ptr;
longword repeated_one;
longword repeated_c;
unsigned char c;
c = (unsigned char) c_in;
/* Handle the first few bytes by reading one byte at a time.
Do this until CHAR_PTR is aligned on a longword boundary. */
for (char_ptr = (const unsigned char *) s;
n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
--n, ++char_ptr)
if (*char_ptr == c)
return (void *) char_ptr;
longword_ptr = (const longword *) char_ptr;
/* All these elucidatory comments refer to 4-byte longwords,
but the theory applies equally well to any size longwords. */
/* Compute auxiliary longword values:
repeated_one is a value which has a 1 in every byte.
repeated_c has c in every byte. */
repeated_one = 0x01010101;
repeated_c = c | (c << 8);
repeated_c |= repeated_c << 16;
if (0xffffffffU < (longword) -1)
{
repeated_one |= repeated_one << 31 << 1;
repeated_c |= repeated_c << 31 << 1;
if (8 < sizeof (longword))
{
size_t i;
for (i = 64; i < sizeof (longword) * 8; i *= 2)
{
repeated_one |= repeated_one << i;
repeated_c |= repeated_c << i;
}
}
}
/* Instead of the traditional loop which tests each byte, we will test a
longword at a time. The tricky part is testing if *any of the four*
bytes in the longword in question are equal to c. We first use an xor
with repeated_c. This reduces the task to testing whether *any of the
four* bytes in longword1 is zero.
We compute tmp =
((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
That is, we perform the following operations:
1. Subtract repeated_one.
2. & ~longword1.
3. & a mask consisting of 0x80 in every byte.
Consider what happens in each byte:
- If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
and step 3 transforms it into 0x80. A carry can also be propagated
to more significant bytes.
- If a byte of longword1 is nonzero, let its lowest 1 bit be at
position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
the byte ends in a single bit of value 0 and k bits of value 1.
After step 2, the result is just k bits of value 1: 2^k - 1. After
step 3, the result is 0. And no carry is produced.
So, if longword1 has only non-zero bytes, tmp is zero.
Whereas if longword1 has a zero byte, call j the position of the least
significant zero byte. Then the result has a zero at positions 0, ...,
j-1 and a 0x80 at position j. We cannot predict the result at the more
significant bytes (positions j+1..3), but it does not matter since we
already have a non-zero bit at position 8*j+7.
So, the test whether any byte in longword1 is zero is equivalent to
testing whether tmp is nonzero. */
while (n >= sizeof (longword))
{
longword longword1 = *longword_ptr ^ repeated_c;
if ((((longword1 - repeated_one) & ~longword1)
& (repeated_one << 7)) != 0)
break;
longword_ptr++;
n -= sizeof (longword);
}
char_ptr = (const unsigned char *) longword_ptr;
/* At this point, we know that either n < sizeof (longword), or one of the
sizeof (longword) bytes starting at char_ptr is == c. On little-endian
machines, we could determine the first such byte without any further
memory accesses, just by looking at the tmp result from the last loop
iteration. But this does not work on big-endian machines. Choose code
that works in both cases. */
for (; n > 0; --n, ++char_ptr)
{
if (*char_ptr == c)
return (void *) char_ptr;
}
return NULL;
}
//mxchipWNET.o
int is_nfc_up(void){
return 0;
}
__weak void *__iar_dlmallinfo(void)
{
return NULL;
}
__weak int aes_decrypt(void)
{
return 0;
}
int nfc_config_stop(void)
{
return 0;
}
int uart_init(void)
{
return 0;
}
int __data_GetMemChunk(void)
{
return 0;
}
char *strdup(const char *src)
{
int len;
char *dst;
if (src == NULL)
return NULL;
if (src[0] == 0)
return NULL;
len = strlen(src) + 1;
dst = (char*)malloc(len);
if (dst)
memcpy(dst, src, len);
return dst;
}
int mico_platform_get_rtc_time (){
return 1;
}
int mico_platform_set_rtc_time (){
return 0;
}
#endif

40
mico-os/platform/RVMDK/memory_debugger.c
View File

@@ -1,20 +1,20 @@
/***
* File: patch_keil.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#if defined(__CC_ARM)
struct mxchip_mallinfo *mxchip_memory_info(void);
struct mxchip_mallinfo* mico_memory_info(void)
{
return mxchip_memory_info();
}
#endif
/***
* File: patch_keil.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#if defined(__CC_ARM)
struct mxchip_mallinfo *mxchip_memory_info(void);
struct mxchip_mallinfo* mico_memory_info(void)
{
return mxchip_memory_info();
}
#endif

62
mico-os/platform/RVMDK/thread_safe.c
View File

@@ -1,31 +1,31 @@
/***
* File: patch_keil.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#include "common.h"
#include "mico_rtos.h"
#ifndef NO_MICO_RTOS
USED int _mutex_initialize(void* mutex)
{
return 1;
}
USED void _mutex_acquire(void* mutex)
{
mico_rtos_suspend_all_thread();
}
USED void _mutex_release(void* mutex)
{
mico_rtos_resume_all_thread();
}
USED void _mutex_free(void* mutex)
{
}
#endif
/***
* File: patch_keil.c
* porting iar project to keil, something need to do.
*
* Created by JerryYu @ Jan 13rd,2015
* Ver: 0.1
* */
#include "common.h"
#include "mico_rtos.h"
#ifndef NO_MICO_RTOS
USED int _mutex_initialize(void* mutex)
{
return 1;
}
USED void _mutex_acquire(void* mutex)
{
mico_rtos_suspend_all_thread();
}
USED void _mutex_release(void* mutex)
{
mico_rtos_resume_all_thread();
}
USED void _mutex_free(void* mutex)
{
}
#endif