Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.1k views
in Technique[技术] by (71.8m points)

c - Unable to pass '#' character as a command-line argument

I can't pass strings starting with # as command-line arguments.

Here is a simple test:

#include <stdio.h>

int main(int argc, char *argv[])
{
    for (int i = 1; i < argc; i++)
        printf("%s ", argv[i]);

    putchar('
');

    return 0;
}

If I input the arguments as follows:

2 4 # 5 6

The value of argc is 3 and not 6. It reads # and stops there. I don't know why, and I can't find the answer in my copies of The C Programming Language and C Primer Plus.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

# begins a comment in Unix shells, much like // in C.

This means that when the shell passes the arguments to the progam, it ignores everything following the #. Escaping it with a backslash or quotes will mean it is treated like the other parameters and the program should work as expected.

2 4 # 5 6

or

2 4 '#' 5 6

or

2 4 "#" 5 6

Note that the # is a comment character only at the start of a word, so this should also work:

2 4#5 6

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...