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我正在编写代码来读取和写入文本文件。但是当我运行我的代码时，我得到了一个 SIGSEGV 运行时错误。我需要帮助找出代码有什么问题。

#include <stdio.h>
#include <stdlib.h>
int main(){
    /*Variable for writing to file */
    char *name[] = {"Raju", "Nayan", "Hanif"};
    char *id[] = {"100", "101", "102"};
    float cgpa[] = {3.50, 5.00, 4.00};
    /*Variable for reading from file*/
    char *getname, *getid;
    float getcgpa;
    int i;
    FILE *fp;
    fp = fopen("f:\\raju.nog", "w"); /*Create new file */
    for(i=0; i<5; i++){
        fprintf(fp, "%-10s%-10s%-1.2f\n", name[i], id[i], cgpa[i]); /*write to file*/
    }
    fclose(fp);
    /*read from file and print to screen*/
    fp = fopen("f:\\raju.nog", "r");
    while(fscanf(fp, "%s %s %f", &getname, &getid, &getcgpa) != EOF){
        printf("%-10s%-10s%-1.2f\n", getname, getid, getcgpa);
    }

    system("PAUSE");
}
4

5 回答 5

7

数组：

char *name[] = {"Raju", "Nayan", "Hanif"};
char *id[] = {"100", "101", "102"};
float cgpa[] = {3.50, 5.00, 4.00};

声明 3 个元素，但

for(i=0; i<5; i++){
    fprintf(fp, "%-10s%-10s%-1.2f\n", name[i], id[i], cgpa[i]); /*write to file*/
}

正在尝试迭代 5 次以上，从而超出了数组的范围。

于 2012-09-14T15:21:30.253 回答
5

您的代码有几个问题：

  • 您正在将每个变量的 5 个元素写入文件，但每个变量只有三个元素 - 导致缓冲区溢出
  • getname, getid 变量只是指针，没有存储空间

要找到这些错误，您应该能够使用调试器单步执行。

这是有效的代码：

#include <stdio.h>
#include <stdlib.h>
int main(){
    /*Variable for writing to file */
    char *name[] = {"Raju", "Nayan", "Hanif"};
    char *id[] = {"100", "101", "102"};
    float cgpa[] = {3.50, 5.00, 4.00};
    /*Variable for reading from file*/

    // need storage space
    char getname[40], getid[40];
    float getcgpa;
    int i;
    FILE *fp;
    fp = fopen("raju.nog", "w"); /*Create new file */

    // Only 3 elements
    for(i=0; i<3; i++){
        fprintf(fp, "%-10s%-10s%-1.2f\n", name[i], id[i], cgpa[i]); /*write to file*/
    }
    fclose(fp);
    /*read from file and print to screen*/
    fp = fopen("raju.nog", "r");
    while(fscanf(fp, "%s %s %f", &getname, &getid, &getcgpa) != EOF){
        printf("%-10s%-10s%-1.2f\n", getname, getid, getcgpa);
    }
    return(0);
}

如果您可以访问 gdb，您可能会发现第一个问题，然后是第二个问题。这是我的 GDB 会话的日志，指出导致代码中第一个错误的行：

(gdb) run
Starting program: /home/jrn/SO/a.out 

Program received signal SIGSEGV, Segmentation fault.
0x005cc35e in vfprintf () from /lib/libc.so.6
(gdb) bt
#0  0x005cc35e in vfprintf () from /lib/libc.so.6
#1  0x005d33cf in fprintf () from /lib/libc.so.6
#2  0x08048565 in main () at sowrong.c:15
(gdb) list
1   #include <stdio.h>
2   #include <stdlib.h>
3   int main(){
4       /*Variable for writing to file */
5       char *name[] = {"Raju", "Nayan", "Hanif"};
6       char *id[] = {"100", "101", "102"};
7       float cgpa[] = {3.50, 5.00, 4.00};
8       /*Variable for reading from file*/
9       char *getname, *getid;
10      float getcgpa;
(gdb) list
11      int i;
12      FILE *fp;
13      fp = fopen("f:\\raju.nog", "w"); /*Create new file */
14      for(i=0; i<5; i++){
15          fprintf(fp, "%-10s%-10s%-1.2f\n", name[i], id[i], cgpa[i]); /*write to file*/
16      }
17      fclose(fp);
18      /*read from file and print to screen*/
19      fp = fopen("f:\\raju.nog", "r");
20      while(fscanf(fp, "%s %s %f", &getname, &getid, &getcgpa) != EOF){
(gdb) list
21          printf("%-10s%-10s%-1.2f\n", getname, getid, getcgpa);
22      }
23  }

注意这一行：#2 0x08048565 in main () at sowrong.c:15
这告诉你第 15 行发生了段错误。之后你可以列出文件以查看该行。

所以现在在修复第 15 行之后，你再次编译你的程序并通过 GDB 运行它来找到第二个问题。这是我的 GDB 会话的日志：

(gdb) run
Starting program: /home/jrn/SO/a.out 

Program received signal SIGSEGV, Segmentation fault.
0x005cc35e in vfprintf () from /lib/libc.so.6
(gdb) bt
#0  0x005cc35e in vfprintf () from /lib/libc.so.6
#1  0x005d3410 in printf () from /lib/libc.so.6
#2  0x080485be in main () at sowrong.c:21
(gdb) list
1   #include <stdio.h>
2   #include <stdlib.h>
3   int main(){
4       /*Variable for writing to file */
5       char *name[] = {"Raju", "Nayan", "Hanif"};
6       char *id[] = {"100", "101", "102"};
7       float cgpa[] = {3.50, 5.00, 4.00};
8       /*Variable for reading from file*/
9       char *getname, *getid;
10      float getcgpa;
(gdb) list
11      int i;
12      FILE *fp;
13      fp = fopen("f:\\raju.nog", "w"); /*Create new file */
14      for(i=0; i<3; i++){
15          fprintf(fp, "%-10s%-10s%-1.2f\n", name[i], id[i], cgpa[i]); /*write to file*/
16      }
17      fclose(fp);
18      /*read from file and print to screen*/
19      fp = fopen("f:\\raju.nog", "r");
20      while(fscanf(fp, "%s %s %f", &getname, &getid, &getcgpa) != EOF){
(gdb) list
21          printf("%-10s%-10s%-1.2f\n", getname, getid, getcgpa);
22      }
23  }

再次注意这一行：#2 0x080485be in main () at sowrong.c:21
第二个 SEGFAULT，因为您的变量只是没有存储空间的指针。

于 2012-09-14T16:22:17.453 回答
3

你没有 malloc 内存

 char *getname, *getid;
于 2012-09-14T15:21:10.380 回答
3

char *getname, *getid;只分配指针，而不是内存来存储字符串。你需要类似的东西：

char getname[11];
char getid[11];

我已经从您的陈述中猜到了大小printf，但您可能想要比这些更大的数组。

于 2012-09-14T15:21:14.497 回答
2

您的 for 循环计数器从 0 变为 4，但您为数组name，id并且cgpa只有 3 个元素：因此您将超出数组边界。

于 2012-09-14T15:22:49.623 回答