如何在脚本bash中传递位置参数？

I start programming in c language, and I ask myself this question.

我开始用c语言编程,我问自己这个问题。

I have a file lol.c :

我有一个文件lol.c:

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

int main (int argc, char *argv[]) {
    printf ("argc\t= %d\n", argc);

    for (int i=0; i < argc; i++)
        printf ("argv[%i]\t= %s\n", i, argv[i]);

    return 0;
}

So when I execute the file (after compilation) :

所以当我执行文件时(编译后):

./lol "abc" "def"

He returns so :

他回来了:

argc    = 3
argv[0] = ./lol
argv[1] = abc
argv[2] = def

Now, I have a bash script lol.sh :

现在,我有一个bash脚本lol.sh:

#!/bin/bash

./lol "abc"

When I execute the file :

当我执行文件时:

bash lol.sh def

So we get :

所以我们得到:

argc    = 2
argv[0] = ./lol
argv[1] = abc

How to retrieve the positional parameters passed to the bash script (from the file lol.c) ?

如何检索传递给bash脚本的位置参数(来自文件lol.c)?

My first solution is to pass the positional parameters to the lol command like this (lol.sh) :

我的第一个解决方案是将位置参数传递给lol命令,如下所示(lol.sh):

#!/bin/bash

./lol "abc" "$*"

He returns so :

他回来了:

argc    = 3
argv[0] = ./lol
argv[1] = abc
argv[2] = def

I am not very satisfied with this solution, is there another way for that ?

我对这个解决方案不是很满意,还有另外一种方法吗?

Thanks

1 个解决方案

#1

While "$*" works, it mashes all the arguments into one, with a space separating the original arguments. You could see this by running:

虽然“$ *”有效,但它将所有参数合并为一个,并用原始参数分隔空格。你可以通过运行看到这个:

bash lol.sh def ghi jkl

which would produce:

会产生:

argc    = 3
argv[0] = ./lol
argv[1] = abc
argv[2] = def ghi jkl

and running bash lol.sh would produce:

并运行bash lol.sh将产生:

argc    = 3
argv[0] = ./lol
argv[1] = abc
argv[2] =

Using "$*" is seldom the correct notation; you should normally use "$@" which relays the arguments verbatim as separate arguments, and handles 'no arguments' correctly (whereas "$*" creates an empty string as an argument).

使用“$ *”很少是正确的表示法;你通常应该使用“$ @”将参数逐字传递为单独的参数,并正确处理“无参数”(而“$ *”创建一个空字符串作为参数)。

This command should do what you want:

这个命令应该做你想要的:

./lol "abc" "$@"

Then the output from bash lol.sh def ghi jkl would be:

那么bash lol.sh def ghi jkl的输出将是:

argc    = 5
argv[0] = ./lol
argv[1] = abc
argv[2] = def
argv[3] = ghi
argv[4] = jkl

and the output from bash lol.sh would be:

并且bash lol.sh的输出将是:

argc    = 2
argv[0] = ./lol
argv[1] = abc

另请参见如何在bash脚本中迭代参数?)

#1