Why isn't argc same as strlen(argv[i])?

Question

Why does my code give segmentation fault when I use strlen(argv[i]) instead of argc?

#include<stdio.h>
#include<ctype.h>
#include<string.h>

int main(int argc , char* argv[])
{
    for(int i=0;i<strlen(argv[i]);i++)
    printf("%s\n",argv[i]);
    return 0;
}

The output is

./a.out
hello
world
Segmentation fault (core dumped)

Because you don't take into account `argc`. The `strlen(argv[i])` doesn't make any sense. — Jabberwocky, Nov 02 '17 at 15:39
You need to be careful when using `strlen()` that doesn't give the count of null terminated byte. — danglingpointer, Nov 02 '17 at 15:43
Would you please tell, what you want to achieve with the above program, then its easy to follow? — danglingpointer, Nov 02 '17 at 15:44
@LethalProgrammer I think he just wants to print the command line arguments, see answer below. — Jabberwocky, Nov 02 '17 at 15:47
`argc` tells you the number of arguments (including the executable name). `argv` is a list of those arguments. If you want to print all the arguments, you must loop over `argc`. Instead, you're looping over the length of `argv[i]`,, as soon as `i` becomes `>= argc`, you're invoking undefined behavior, manifested in a segfault — yano, Nov 02 '17 at 15:47

score 2 · Accepted Answer · answered Nov 02 '17 at 15:45

2

Use argc:

int main(int argc , char* argv[])
{
    for(int i = 0; i < argc; i++)
        printf("%s\n", argv[i]);
    return 0;
}

For more information, see the answer to this question: What does int argc, char *argv[] mean?

answered Nov 02 '17 at 15:45

Andrew Cottrell

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Why isn't argc same as strlen(argv[i])?

1 Answers1