在C中返回分配的指针

Question

I have the following program:

#include <stdio.h>
#include <stdlib.h>

char* getStr(int length) {
    char* chars = malloc(length + 1);
    int i;
    for(i = 0; i < length; i++)
        chars[i] = 'X';
    chars[i] = '\0';
    // no call to free()
    return chars;
}

int main(int argc, char** argv) {
    char* str;
    str = getStr(10);
    printf("%s\n", str);
    free(str);
    return EXIT_SUCCESS;
}

It prints 10 X 's, as I expected. Would it behave like this on any platform with any compiler? Is the memory still malloced after getStr() returns?

(I don't want to pass the pointer as argument :D)

Answer 1

如果您使用malloc分配内存，那么它将一直保持分配状态，直到您显式调用free为止，无论它在函数之间如何传递，返回等。

Answer 2

Yes, the code looks valid and the behavior should be reliable with any C compiler.

And, yes, the memory is still allocated after getStr() returns. So the call to free() is also correct.

Don't forget to check if malloc() returns NULL , in the event there is insufficient memory.