Reading string character by character in C

Question

So I have a string passed into main function: int main(int argc, char* argv[])

I understand argc (which is 2 in this case), but don't understand how I can read argv[] character by character? When I print argv[0] shouldn't that print the first character in the array of characters for that string?

Thanks

The values passed on the command line start with argv[1]. The first character of that would be argv[1][0]. — Retired Ninja
– Retired Ninja, Commented Sep 27, 2014 at 7:22
argv[] is an array of strings(null terminated character arrays). Thus argv[0] gives the first string. To get the first character of the first string use *argv[0] or argv[0][0]. — Abhishek Choubey
– Abhishek Choubey, Commented Sep 27, 2014 at 8:18

BLUEPIXY · Accepted Answer · 2014-09-27 07:28:26Z

7

sample

#include <stdio.h> 

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

answered Sep 27, 2014 at 7:28

BLUEPIXY

40.2k7 gold badges37 silver badges73 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

aNobody Over a year ago

Thanks, I swear I tried the same thing but it didn't work. Does ++i vs i++ make any difference? I couldn't understand other people's explanations.

Darren Stone Over a year ago

Please share the code that didn't work. No, in this context the pre or post increment operator makes no difference. It does make a difference if it is part of an expression being assigned or evaluated, as opposed to an increment operation alone.

BLUEPIXY Over a year ago

@aNobody You should post together the code that did not work well when you asked question.

BLUEPIXY Over a year ago

@aNobody Does ++i vs i++ make any difference? : There is no much difference if it's running alone.

Bob Shaffer Over a year ago

@aNobody ++i is more efficient by a very tiny amount because the additional storage for the return value need not be allocated.

ani627 · Accepted Answer · 2014-09-27 07:44:22Z

0

One more example:

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

int main (int argc, char *argv[])
{
    if(argc != 2)
    {
        //argv[0] is name of executable
        printf("Usage: %s argument\n", argv[0]);
        exit(1);
    }
    else
    {
        int i;
        int length = strlen(argv[1]);
        for(i=0;i<length;i++)
        {
            printf("%c",argv[1][i]);
        }
        printf("\n");
        return 0;
    }
}

answered Sep 27, 2014 at 7:44

ani627

6,0958 gold badges42 silver badges48 bronze badges

Collectives™ on Stack Overflow

Reading string character by character in C

2 Answers 2

5 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related