C - Command line arguments as strings?

Question

So I'm confused on how exactly command line arguments work in C... so I have these command line arguments that I'm giving:

 ./myclient1 MyPCName 12 7894

So I want to read argv[2] (12) as a String... but I'm confused exactly how values are stored in the command line. I looked at both this SO post,this this link but I'm still confused... what is the data type of argv[2]? Is it an integer? Or are all command line arguments originally strings? So argv[2] is actually:

  argv: 
  [0]
  [1]
  [2] --> 1 | 2 | \0

I'm just really confused.... currently, I just converted to an integer using atoi() and then converted back to a string using snprintf, but it's not working correctly and I'm wondering if I need to do this at all.

I'm new to C so any help would be greatly appreciated, thanks!!

[edit]

this is what I had done before:

int main(int argc, char *argv[])
{
 clientID = atoi(argv[2]);
snprintf(clibuff,300,"%d",clientID);  //now clibuff has the value of 
                                      //clientID in a string.
}

Please post some code of what you've tried.

QuestionC
– QuestionC

2015-09-28 20:38:33 +00:00
Commented Sep 28, 2015 at 20:38 — QuestionC
– QuestionC, Commented Sep 28, 2015 at 20:38
All command line arguments come in as strings.

QuestionC
– QuestionC

2015-09-28 20:39:05 +00:00
Commented Sep 28, 2015 at 20:39 — QuestionC
– QuestionC, Commented Sep 28, 2015 at 20:39

David Ranieri · Accepted Answer · 2015-09-28 20:40:09Z

3

Command line arguments are stored as an array of zero-terminated strings.

So to answer your question, the type of argv[2] is char *

The type of argv is char **

edited Sep 28, 2015 at 20:40

David Ranieri

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answered Sep 28, 2015 at 20:39

Matthew Logan

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1 Comment

ocean800 Over a year ago

Oh... thanks I see. So then my understanding was largely correct. So then 12 is stored as a string with an array of char....

SalientGreen · Accepted Answer · 2015-09-28 20:40:53Z

2

Here are some great explanations on CLI arguments:

http://www.tutorialspoint.com/cprogramming/c_command_line_arguments.htm

https://en.wikibooks.org/wiki/A_Little_C_Primer/C_Command_Line_Arguments

answered Sep 28, 2015 at 20:40

SalientGreen

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Comments

John Bollinger · Accepted Answer · 2015-09-28 20:45:28Z

1

A hosted C implementation (such as you surely are using) permits exactly two signatures for main(), and if you want access to the command line arguments then only one will do:

int main(int argc, char *argv[]);

You can spell it in couple of other, equivalent ways, but I prefer that one because it emphasizes the array nature of argv. Each element, of course, points to one of the first character of a standard C string containing the corresponding command-line argument. Thus, the type of argv[2] is always char *, and you don't need to do anything special to handle it as a C string.

answered Sep 28, 2015 at 20:45

John Bollinger

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1 Comment

ocean800 Over a year ago

thanks... I think this answer made things much more clearer.

Tom · Accepted Answer · 2015-09-28 20:46:57Z

1

a copy of your test code would be useful..

I usually declare argv this way:

int main(int argc, char **argv);

When the program starts, the arguments to main will have been initialized to this:

argc is greater than zero.
argv[argc] is a null pointer.
argv[0] to argv[argc-1] are pointers to strings
argv[0] will be a string containing the program's name or a null string if that is not available. Remaining elements of argv represent the arguments supplied to the program.

This Testprogram should work:

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

int main(int argc, char **argv)
{
        while(argc--)
                printf("%s\n", *argv++);
        exit(EXIT_SUCCESS);
}

answered Sep 28, 2015 at 20:46

Tom

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Comments

Sebas · Accepted Answer · 2015-09-28 20:48:53Z

0

The argv[0] is always the name of your program, and yes all your arguments are string you should count your arguments since the argv[1].

answered Sep 28, 2015 at 20:48

Sebas

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