访问冲突读取位置0x00000000。与argv []

时间:2013-06-06 04:03:35

标签: c pointers access-violation argv strcmp

我正在运行以下程序并遇到错误:

First-chance exception at 0x0f32d440 (msvcr100d.dll) in c.exe: 0xC0000005: Access violation reading location 0x00000000.
Unhandled exception at 0x772815de in c.exe: 0xC0000005: Access violation reading location 0x00000000.
The program '[9048] c.exe: Native' has exited with code -1073741510 (0xc000013a).

这是代码

#include <string.h>
#include <stdio.h> 
#include <stdlib.h> 
int main(int argc, char *argv[], char *env[]) //char *argv[] 
{ 
int i; 

printf("These are the %d command- line arguments passed to main:\n\n", argc); 
if(strcmp(argv[1],"123")==0)                
    {
        printf("success\n");
     }
else
for(i=0; i<=argc; i++) 
    //if(strcmp(argv[1],"abc")==0)

    printf("argv[%d]:%s\n", i, argv[i]);
/*printf("\nThe environment string(s)on this system are:\n\n"); 
for(i=0; env[i]!=NULL; i++) 
printf(" env[%d]:%s\n", i, env[i]);*/ 
system("pause");
} 

问题应该是strcmp函数,但我不知道如何解决它。 有人可以帮忙吗?

2 个答案:

答案 0 :(得分:3)

你(至少)有两个问题。

首先是这样做:

if(strcmp(argv[1],"123")==0) 

没有先检查argc >= 2

第二个是:

for(i=0; i<=argc; i++) 

因为你应该处理0argc - 1的{​​{1}}参数。该循环的作用是流程参数0argcargv[argc]始终为NULL

以下程序说明了解决此问题的一种方法:

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

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

    printf ("These are the %d command-line argument(s) passed to main:\n", argc);
    if ((argc >= 2) && (strcmp (argv[1], "123") == 0)) {
        printf ("   success\n");
    } else {
        for (i = 0; i < argc; i++)  {
            printf ("   argv[%d] = [%s]\n", i, argv[i]);
        }
    }
    return 0;
}

您可以看到只有在确保正确填充"123"后才能与argv[1]进行比较。此外,循环已更改为排除argv[argc],因为这不是参数之一。成绩单如下:

pax> testprog
These are the 1 command-line argument(s) passed to main:
   argv[0] = [testprog]

pax> testprog 123
These are the 2 command-line argument(s) passed to main:
   success

pax> testprog a b c
These are the 4 command-line argument(s) passed to main:
   argv[0] = [testprog]
   argv[1] = [a]
   argv[2] = [b]
   argv[3] = [c]

答案 1 :(得分:1)

for(i=0; i<=argc; i++)应为for(i=0; i<argc; i++)

C / C ++数组为0到n-1。你在阵列末尾跑了1个点。