C语言 Malloc 返回错误

Question

分配数组并释放它后，我显示一条错误消息，如下所示：

*** glibc detected *** ./Q3: malloc(): memory corruption (fast): 0x09a092f8 ***

我该如何补救这种情况？其背后的潜在结果可能是什么？

使用的代码如下：

// The compilation command used is given below
// gcc Q3.c nrutil.c DFRIDR.c -lm -o Q3
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include "nr.h"
#define LIM1 20.0
#define a -5.0
#define b 5.0
#define pre 100.0 // This defines the pre
/* This file calculates the func at given points, makes a 
 * plot. It also calculates the maximum and minimum of the func
 * at given points and its first and second numerical derivative.
*/
float func(float x)
{
    return exp(x / 2) / pow(x, 2);
}
int main(void)
{
    FILE *fp = fopen("Q3data.dat", "w+"), *fp2 = fopen("Q3results.dat", "w+");
    int i; // Declaring our loop variable
    float x, y, min, max, err, nd1, nd2;
    // These arrays are defined so that we can pass them into Numerical Recipes routines
    float * xp = malloc(((b - a) / pre) * sizeof(float));
    float * yp = malloc(((b - a) / pre) * sizeof(float));
    float yp1 = dfridr(func, a, 0.1, &err), ypn = dfridr(func, b, 0.1, &err);
    // Define the initial value of the func to be the minimum
    min = func(0); 
    for(i = 0; x < LIM1 ; i++)
    {   
        x = i / pre; // There is a singularity at x = 0 
        y = func(x);
        if(y < min)
            min = y;
        fprintf(fp, "%f t %f n", x, y);
    }
    fprintf(fp, "nn");
    max = 0;
    for(i = 0, x = a; x < b; i++)
    {   
        xp[i] = a + i / pre;
        yp[i] = func(xp[i]);
        nd1 = dfridr(func, xp[i], 0.1, &err); 
        //nd2 = dfridr((*func), x, 0.1, &err);
        fprintf(fp, "%f t %f t %f t n", xp[i], yp[i], nd1);
        if(y > max)
            max = y;
    }
    free((void *)xp);
    free((void *)yp);
    fprintf(fp2, "The minimum value of f(x) is %f when x is between 0 and 20. n", min);
    fprintf(fp2, "The maximum value of f(x) is %f when x is between -5 and 5. n", max);
    fclose((void *)fp);
    fclose((void *)fp2);
    return 0;
}

修改后的指针 xp、yp 如下：

float * xp = malloc(((b - a) * pre + 1) * sizeof(float));
float * yp = malloc(((b - a) * pre + 1) * sizeof(float));

经过这些修改后，问题仍然存在。

整个代码中的最新修改，但问题仍然存在：

// The compilation command used is given below
// gcc Q3.c nrutil.c DFRIDR.c -lm -o Q3
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include "nr.h"
#define LIM1 20.0
#define a -5.0
#define b 5.0
#define pre 100.0 // This defines the pre
/* This file calculates the func at given points, makes a 
 * plot. It also calculates the maximum and minimum of the func
 * at given points and its first and second numerical derivative.
*/
float func(float x)
{
    return exp(x / 2) / pow(x, 2);
}
int main(void)
{
    FILE *fp = fopen("Q3data.dat", "w+"), *fp2 = fopen("Q3results.dat", "w+");
    int i; // Declaring our loop variable
    float min, max, err, nd1, nd2;
    // These arrays are defined so that we can pass them into Numerical Recipes routines
    //printf("%d n", (int)((b - a) * pre + 1));
    float * x = malloc((int)(LIM1 * pre) * sizeof(float));
    float * y = malloc((int)(LIM1 * pre) * sizeof(float));
    float * xp = malloc((int)((b - a) * pre + 1)* sizeof(float));
    float * yp = malloc((int)((b - a) * pre + 1)* sizeof(float));
    if (xp == 0 || yp == 0 || x == 0 || y == 0)
        {
            printf("ERROR: Out of memoryn");
            return 1;
        }
    float yp1 = dfridr(func, a, 0.1, &err), ypn = dfridr(func, b, 0.1, &err);
    // Define the initial value of the func to be the minimum
    min = func(0); 
    for(i = 0; x[i] < LIM1 ; i++)
    {   
        x[i] = i / pre; // There is a singularity at x = 0 
        y[i] = func(x[i]);
        if(y[i] < min)
            min = y[i];
        fprintf(fp, "%f t %f n", x[i], y[i]);
    }
    fprintf(fp, "nn");
    max = 0;
    for(i = 0; xp[i] < 5.0; i++)
    {   
        xp[i] = a + i / pre;
        yp[i] = func(xp[i]);
        nd1 = dfridr(func, xp[i], 0.1, &err); 
        fprintf(fp, "%f t %f t %f t n", xp[i], yp[i], nd1);
        if(yp[i] > max)
            max = yp[i];
    }
    free((void *)x);
    free((void *)y);
    free((void *)xp);
    free((void *)yp);
    fprintf(fp2, "The minimum value of f(x) is %f when x is between 0 and 20. n", min);
    fprintf(fp2, "The maximum value of f(x) is %f when x is between -5 and 5. n", max);
    fclose(fp);
    fclose(fp2);
    return 0;
}

Answer 1

这个循环永远不会结束，因为x和b永远不会被修改：

for(i = 0, x = a; x < b; i++)
{   
    xp[i] = a + i / pre;
    yp[i] = func(xp[i]);
    nd1 = dfridr(func, xp[i], 0.1, &err); 
    //nd2 = dfridr((*func), x, 0.1, &err);
    fprintf(fp, "%f t %f t %f t n", xp[i], yp[i], nd1);
    if(y > max)
        max = y;
}

因此，即使使用更新的xp和yp指针，在某个时间点，您的代码开始写入超出您分配的内存限制，因此您将获得未定义的行为。您很幸运glibc检测到了这一点（您可能破坏了它的一些内部数据结构）。