本文介绍了的malloc（0）的实际工作？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
Why does malloc(0) actually return a valid pointer for writing ?
char *str = NULL;
str = (char*)malloc(0); // allocate 0 bytes ?
printf("Pointer of str: %p\n", str);
strcpy(str, "A very long string ...................");
printf("Value of str: %s", str);
free(str); // Causes crash if str is too long
Output:
Pointer of str: 0xa9d010
Aborted
Value of str: A very long string ...................
When str is shorter then it just works as it should.
BTW: For compiling I used GCC with "-D_FORTIY_SOURCE=0 -fno-stack-protector"
*** glibc detected *** ..: free(): invalid next size (fast): 0x0000000000a9d010 ***
解决方案
It doesn't return a valid pointer for writing. It returns a valid pointer for not using it. Or it may return NULL as well since the C standard specifies this case to be implementation defined.
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