问题描述
我试图将参数传递给ls命令的execl()函数.但是当我通过
I am trying to pass arguments to execl() function for ls command. But when I pass
/bin/ls -l -a
作为我程序的参数,execl()函数无法识别最后两个参数.为什么呢?代码如下:
as arguments to my program, the execl() function doesn't recognise the last two arguments. Why is that?Here is the code:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int i,childpid;
if(argc!=4)
{
printf("you didn't provide any commandline arguments!\n");
return 0;
}
childpid=fork();
if(childpid<0)
{
printf("fork() failed\n");
return 0;
}
else
if(childpid==0)
{
printf("My ID %d\n\n",getpid());
execl(argv[1],argv[2], argv[3],NULL);
printf("If you can see this message be ware that Exec() failed!\n");
}
while(wait(NULL)>0);
printf("My ID %d, Parent ID %d, CHild ID %d\n", getpid(),getppid(),childpid);
return 0;
}
我在Ubuntu上.
致谢
推荐答案
在运行程序时,/bin/ls
似乎忽略了-l
参数,因为它是在为argv[0]
保留的位置传递的,通常是(相当无用的)程序名称.
When running your program /bin/ls
appears to ignore the -l
argument because it is passed in the position reserved for argv[0]
, which is normally the (rather useless) program name.
具体地说,execl
的第一个参数是要运行的程序,其余参数按原样复制到argv
向量.由于期望argv[0]
包含程序名称,并且实际参数以argv[1]
开头,因此您必须通过两次提供程序名称来补偿:
Specifically, the first argument to execl
is the program to run, and the remaining arguments get copied to argv
vector as-is. Since argv[0]
is expected to contain the program name and the actual arguments begin from argv[1]
, you must compensate by providing the program name twice:
execl(argv[1], argv[1], argv[2], argv[3], (char *) NULL);
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