fread() 中的段错误

发布于 2024-10-25 08:33:47 字数 1407 浏览 6 评论 0原文

我正在尝试使用 C 读取 BMP 图像（灰度），将值保存到数组中，并将该数组转换为值以逗号分隔的字符串。

我的程序在 Windows 7 64 位下运行良好，但由于库兼容性问题，我不得不迁移到 Windows XP 32 位。

我有 1,750 张图像要读取，我想将它们全部存储在一个字符串中。

当我启动程序时，直到第 509 个图像为止一切正常，然后我收到由 fread() 引起的分段错误。这是我的代码：

int i=0,j,k,num,len,length,l;
unsigned char *Buffer;
FILE *fp;
char *string,*finalstring;
char *query;
char tmp2[5],tmp[3];
query = (char *)malloc(sizeof(char)*200000000);
string = (char *)malloc(sizeof(char)*101376);
Buffer = (unsigned char *)malloc(sizeof(unsigned char)*26368);
 BITMAPFILEHEADER bMapFileHeader;
BITMAPINFOHEADER bMapInfoHeader;
length = 0;


  for (k =1;k<1751;k++)
    {
    strcpy(link,"imagepath");
    //here just indexing the images from 0000 to 1750
    sprintf(tmp2,"%.4d",k);
    strcat(link,tmp2);
    strcat(link,".bmp");


       fp = fopen(link, "rb");
       num = fread(&bMapFileHeader,sizeof(BITMAPFILEHEADER),1,fp);
       num = fread(&bMapInfoHeader,sizeof(BITMAPINFOHEADER),1,fp);
    //seek beginning of data in bitmap
     fseek(fp,54,SEEK_SET);
    //read in bitmap file to data

    fread(Buffer,26368,1,fp);
    l=0;

    for(i=1024;i<26368;i++)
    {
      itoa(Buffer[i],tmp,10);
      len = strlen(tmp);
      memcpy(string+l,tmp,len);
      memcpy(string+l+len,",",1);
      l = l+len+1;

    }

    memcpy(query,"",1);
    memcpy(string,"",1);
    printf("%i\n",k);

    }

谢谢

原文

I'm trying to read a BMP image (greyscales) with C, save values into an array, and convert this array to a string with values separated with a comma.

My program worked well under Windows 7 64-bit, but I had to move to Windows XP 32-bit because of library compatibility problems.

I have 1,750 images to read, and I want to store all of them in a single string.

When I launch my program it goes fine until the 509:th image, then I get a Segmentation Fault caused by fread(). Here's my code:

int i=0,j,k,num,len,length,l;
unsigned char *Buffer;
FILE *fp;
char *string,*finalstring;
char *query;
char tmp2[5],tmp[3];
query = (char *)malloc(sizeof(char)*200000000);
string = (char *)malloc(sizeof(char)*101376);
Buffer = (unsigned char *)malloc(sizeof(unsigned char)*26368);
 BITMAPFILEHEADER bMapFileHeader;
BITMAPINFOHEADER bMapInfoHeader;
length = 0;


  for (k =1;k<1751;k++)
    {
    strcpy(link,"imagepath");
    //here just indexing the images from 0000 to 1750
    sprintf(tmp2,"%.4d",k);
    strcat(link,tmp2);
    strcat(link,".bmp");


       fp = fopen(link, "rb");
       num = fread(&bMapFileHeader,sizeof(BITMAPFILEHEADER),1,fp);
       num = fread(&bMapInfoHeader,sizeof(BITMAPINFOHEADER),1,fp);
    //seek beginning of data in bitmap
     fseek(fp,54,SEEK_SET);
    //read in bitmap file to data

    fread(Buffer,26368,1,fp);
    l=0;

    for(i=1024;i<26368;i++)
    {
      itoa(Buffer[i],tmp,10);
      len = strlen(tmp);
      memcpy(string+l,tmp,len);
      memcpy(string+l+len,",",1);
      l = l+len+1;

    }

    memcpy(query,"",1);
    memcpy(string,"",1);
    printf("%i\n",k);

    }

Thanks

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情深已缘浅 2024-11-01 08:33:48

摆脱 malloc 调用中的强制转换。和#include。

在您使用的 64 位和 32 位系统中，该程序的输出是什么？

#include <stdio.h>
int main(void) {
    printf("sizeof (int) is %d\n", (int)(sizeof (int)));
    printf("sizeof (int*) is %d\n", (int)(sizeof (int*)));
    return 0;
}

Get rid of the casts in malloc calls. And #include <stdlib.h>.

What's the output of this program, in both the 64-bit and 32-bit systems you're using?

#include <stdio.h>
int main(void) {
    printf("sizeof (int) is %d\n", (int)(sizeof (int)));
    printf("sizeof (int*) is %d\n", (int)(sizeof (int*)));
    return 0;
}

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