Question

运行c程序时出现以下错误：

*** glibc detected *** ./a.out: double free or corruption (!prev): 0x080b8008 ***

我相信这是因为在程序结束时调用了free（），但我无法弄清楚malloc内存在此之前被释放的位置。这是代码：

#include <stdio.h>
#include <stdlib.h> //malloc
#include <math.h>  //sine

#define TIME 255
#define HARM 32

int main (void) {
    double sineRads;
    double sine;
    int tcount = 0;
    int hcount = 0;
    /* allocate some heap memory for the large array of waveform data */
    double *ptr = malloc(sizeof(double *) * TIME);
    if (NULL == ptr) {
        printf("ERROR: couldn't allocate waveform memory!\n");
    } else {
        /*evaluate and add harmonic amplitudes for each time step */
        for(tcount = 0; tcount <= TIME; tcount++){
            for(hcount = 0; hcount <= HARM; hcount++){
                sineRads = ((double)tcount / (double)TIME) * (2*M_PI); //angular frequency
                sineRads *= (hcount + 1); //scale frequency by harmonic number
                sine = sin(sineRads); 
                *(ptr+tcount) += sine; //add to other results for this time step
            }
        }
        free(ptr);
        ptr = NULL;     
    }
    return 0;
}

编译时使用：

gcc -Wall -g -lm test.c

Valgrind的：

valgrind --leak-check=yes ./a.out

给出：

    ==3028== Memcheck, a memory error detector
==3028== Copyright (C) 2002-2009, and GNU GPL'd, by Julian Seward et al.
==3028== Using Valgrind-3.6.0.SVN-Debian and LibVEX; rerun with -h for copyright info
==3028== Command: ./a.out
==3028== 
==3028== Invalid read of size 8
==3028==    at 0x8048580: main (test.c:25)
==3028==  Address 0x41ca420 is 1,016 bytes inside a block of size 1,020 alloc'd
==3028==    at 0x4024F20: malloc (vg_replace_malloc.c:236)
==3028==    by 0x80484F8: main (test.c:15)
==3028== 
==3028== Invalid write of size 8
==3028==    at 0x8048586: main (test.c:25)
==3028==  Address 0x41ca420 is 1,016 bytes inside a block of size 1,020 alloc'd
==3028==    at 0x4024F20: malloc (vg_replace_malloc.c:236)
==3028==    by 0x80484F8: main (test.c:15)
==3028== 
==3028== 
==3028== HEAP SUMMARY:
==3028==     in use at exit: 0 bytes in 0 blocks
==3028==   total heap usage: 1 allocs, 1 frees, 1,020 bytes allocated
==3028== 
==3028== All heap blocks were freed -- no leaks are possible
==3028== 
==3028== For counts of detected and suppressed errors, rerun with: -v
==3028== ERROR SUMMARY: 8514 errors from 2 contexts (suppressed: 14 from 7)

我对没有自动管理自己内存的语言没有太多经验（因此这项练习在c中学习了一下）但是我被卡住了。任何帮助将不胜感激。

该代码应该是附加音频合成器的一部分。在这方面，它确实有效，并提供存储在ptr。

中的正确输出

感谢。

Answer 1

double *ptr = malloc(sizeof(double *) * TIME);
/* ... */
for(tcount = 0; tcount <= TIME; tcount++)
                         ^^

你超越了阵列。将<=更改为<或分配 SIZE + 1元素
您的malloc错了，您需要sizeof(double)而不是sizeof(double *) ouah
作为*(ptr+tcount)评论，虽然与您的腐败问题没有直接关联，但您使用ptr[tcount]而未初始化

就像样式注释一样，您可能希望使用*(ptr + tcount)代替malloc
您不需要free + SIZE，因为您已经知道{{1}}

Answer 2

更改此行

double *ptr = malloc(sizeof(double *) * TIME);

到

double *ptr = malloc(sizeof(double) * TIME);

Answer 3

1 - 你的malloc（）错了。
2 - 您超出了已分配内存的范围 3 - 您应该初始化分配的内存

以下是需要进行所有更改的程序。我编译并运行......没有错误或警告。

#include <stdio.h>
#include <stdlib.h> //malloc
#include <math.h>  //sine
#include <string.h>

#define TIME 255
#define HARM 32

int main (void) {
    double sineRads;
    double sine;
    int tcount = 0;
    int hcount = 0;
    /* allocate some heap memory for the large array of waveform data */
    double *ptr = malloc(sizeof(double) * TIME);
     //memset( ptr, 0x00, sizeof(double) * TIME);  may not always set double to 0
    for( tcount = 0; tcount < TIME; tcount++ )
    {
         ptr[tcount] = 0; 
    }

    tcount = 0;
    if (NULL == ptr) {
        printf("ERROR: couldn't allocate waveform memory!\n");
    } else {
        /*evaluate and add harmonic amplitudes for each time step */
        for(tcount = 0; tcount < TIME; tcount++){
            for(hcount = 0; hcount <= HARM; hcount++){
                sineRads = ((double)tcount / (double)TIME) * (2*M_PI); //angular frequency
                sineRads *= (hcount + 1); //scale frequency by harmonic number
                sine = sin(sineRads); 
                ptr[tcount] += sine; //add to other results for this time step
            }
        }
        free(ptr);
        ptr = NULL;     
    }
    return 0;
}

Answer 4

我没有检查所有代码，但我的猜测是错误发生在malloc调用中。你必须替换

 double *ptr = malloc(sizeof(double*) * TIME);

代表

 double *ptr = malloc(sizeof(double) * TIME);

因为你想为double赋一个大小（不是指向double的指针的大小）。

c程序中的双重免费或损坏（！prev）错误

4 个答案: