[英]Detecting no option with getopt in C (in Linux)
I want to write a simple C program using terminal in Linux. 我想在Linux中使用终端编写一个简单的C程序。 I don't know how to check if no option was provided during program executing:
我不知道如何检查程序执行期间是否未提供任何选项:
./program.a
Here's my script: 这是我的脚本:
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv)
{
int opt;
while ((opt = getopt (argc, argv, "il:")) != -1)
switch (opt)
{
case 'i':
printf("This is option i");
break;
case 'l':
printf("This is option l");
break;
default:
fprintf(stderr,"Usage: %s [-i] opt [-l] opt\n",argv[0]);
}
if (argc == -1) {
printf("Without option");
}
}
So the output with: 因此输出结果如下:
./program.a
Should be: 应该:
"Without option"
I tried to do it with "if" and set argc to -1, 0 or NULL, but it doesn't work. 我尝试使用“ if”将argc设置为-1、0或NULL,但是它不起作用。 I know that in bash I can use sth like that: if [ $# -eq 0] or if [-z "${p}" ], to check if no option was provided, but in CI have no idea how to check that...
我知道在bash中我可以这样使用sth:if [$#-eq 0]或if [-z“ $ {p}”],检查是否未提供任何选项,但在CI中不知道如何检查那...
I have a second question too: is it possible to somehow combine bash functions with C code in one script/program? 我也有第二个问题:是否可以在一个脚本/程序中以某种方式将bash函数与C代码结合起来?
Thanks for any hints. 感谢您的任何提示。 B
乙
我认为如果不传递任何参数(选项)来启动,则需要检查argc编号,您的argc将为1,否则为1 + count(选项)。
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