Valgrind显示分配的内存比实际多

Question

我在C中编写了一些简单的代码来测试一些内存分配和指针：

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


int *randomAlloc(int n) {
    int *address = NULL, i = 0;

    address = malloc (n * sizeof(int));
    for (i = 0; i < n ; i++){
        *(address + i) = i ;
    }
    return address;

}

int main(int argc, char* argv[] ) {

    int *address;
    int n;
    printf("Type vector size: ");
    scanf("%d", &n);
    address = randomAlloc(n);

    free(address);
}

但是由于某些原因，当我输入4作为输入valgrind输出时：

==2375== Memcheck, a memory error detector
==2375== Copyright (C) 2002-2015, and GNU GPL'd, by Julian Seward et al.
==2375== Using Valgrind-3.12.0 and LibVEX; rerun with -h for copyright info
==2375== Command: ./a.out
==2375== 
Type vector size: 4
==2375== 
==2375== HEAP SUMMARY:
==2375==     in use at exit: 0 bytes in 0 blocks
==2375==   total heap usage: 3 allocs, 3 frees, 2,064 bytes allocated
==2375== 
==2375== All heap blocks were freed -- no leaks are possible
==2375== 
==2375== For counts of detected and suppressed errors, rerun with: -v
==2375== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

代码中只有一个alloc和一个空闲。由于n = 4，我希望它分配4 * 4（sizeof（int））= 16个字节。这是哪里来的？

Answer 1

Valgrind会跟踪应用程序中发生的所有内存分配，包括由C库在内部进行的分配。它不（也不能）限于您显式进行的分配，因为C库可以返回指向其内部分配的内存的指针。

许多标准I / O实现都会分配缓冲区，以供printf()和/或使用scanf()，这可能是您看到的数字的原因。