Thread: C User input

I'm struggling with the whole pointer concept at the moment. For example I have a piece of code for testing the first char of a command line arguement but I can't get it to work.

I have tried different ways but at the minute I have

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
	char input = argv[1];
	printf("\n\nInput : %s\n\n", input);
	if(input[0]=="M")
	{
		printf("\nThis is a message\n");
	}
	else if(input[0] == "L")
	{
		printf("\nThis is a log-in\n");
	}
	else
	{
		printf("\n\nwe have no match\n\n");
	}
}

Could someone try and explain to me where I'm going wrong? This version says :

prefix.c: In function ‘main’:
prefix.c:5: warning: initialisation makes integer from pointer without a cast
prefix.c:7: error: subscripted value is neither array nor pointer
prefix.c:11: error: subscripted value is neither array nor pointer

Thanks!

char input = argv[1];

argv[1] is an array of char (according to the definition above "int main(int argc, char *argv[])"

So, you've almost got it, now you simply need to subscript that array as well.

Edit:

So, after reading your code more, I would make char input as a pointer, eg char *input = ...

Then, if you know this syntax, use a switch statement in place of the nested if else -- though this is not a requirement (in fact it adds some overhead for the processing and the if-else is better), but it is more readable.

Thanks for reply. However,

I tried doing

Code:

char input = argv[1];

and i got :

prefix.c: In function ‘main’:
prefix.c:7: error: subscripted value is neither array nor pointer
prefix.c:11: error: subscripted value is neither array nor pointer

Originally Posted by quzah

'argv' is an array of pointers to character. That means that 'argv[x]' is a pointer to a character. That means that you need to dereference THAT to get a single character.


Quzah.

Is there an echo in this thread???

I did read it. However if you read my post you'd see I said I was new to C and struggling with the concept and usage of pointers.

Originally Posted by Martin_T

I'm struggling with the whole pointer concept at the moment. For example I have a piece of code for testing the first char of a command line arguement but I can't get it to work.

I have tried different ways but at the minute I have

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
	char input = argv[1];
	printf("\n\nInput : %s\n\n", input);
	if(input[0]=="M")
	{
		printf("\nThis is a message\n");
	}
	else if(input[0] == "L")
	{
		printf("\nThis is a log-in\n");
	}
	else
	{
		printf("\n\nwe have no match\n\n");
	}
}

Could someone try and explain to me where I'm going wrong? This version says :



Thanks!

You are having some problems in your code

first of all argv[0], argv[1] are pointer to char array means you need a char array to store argv[0] or argv[1] and so on....

like

Code:

char input[100];
strcpy( input, argv[0]); // copy the whole argv[0] string to input string

and other one is if you are comparing string like input with "M" as i think you know in C/C++ string means double quotes " " and a single char means ' ' single quotes.

So you should use strcmp to compare the strings

Code:

if (strcmp(input, "M") == 0) {
  // true
} else {
  // false
}

In case anyone has the same issue.

In c you cannot directly compare the value of 2 strings in a condition like if(string1==string2)

Are you asking a question or giving suggestion to forum members ??

Sorry yes, I was just wanting to post what I found in case anyone was searching and came across the same problem. I have solved the issue.

Dude in C you cannot do that this is simple string comparison with == operator

I'm confused. If your saying you can't use "==" for string comparison in C then we agree. I had to use strcmp to get my code to work.