Need help with displaying argc and argv

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I run the following code on Linux:

C++

#include <stdio.h>
#include <string.h>
int main( int argc, char ** argv )
{
    int I_count;
    for( I_count = 0; I_count < argc; I_count++ )
    {
            fputs( argv[I_count], stdout );
    }
    putchar( '\n' );
    return 0;
}

And get this result in terminal window:
Input: ./1 jiawei

Output: ./jiawei

I don't understand why, if you know please answer me. Thanks.

Posted 24-May-10 23:29pm

harryxiyou

Updated 25-May-10 0:07am

Moak

v3

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CPallini · Answer 1 · 2010-05-25T00:22:00

Solution 1

I suppose the program output is actually:

C

./1jiawei

since you didn't put any separator between the printed arguments.
Try

C

...
    for( I_count = 0; I_count < argc; I_count++ )
    {
            fputs( argv[I_count], stdout );
            fputs( " ", stdout);
    }
...

:)

Posted 25-May-10 0:22am

CPallini