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?
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0
至argc - 1
的{{1}}参数。该循环的作用是流程参数0
到argc
,argv[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。