C User input

This is a discussion on C User input within the C Programming forums, part of the General Programming Boards category; I'm struggling with the whole pointer concept at the moment. For example I have a piece of code for testing ...

  1. #1
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    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!

  2. #2
    {Jaxom,Imriel,Liam}'s Dad Kennedy's Avatar
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    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.

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    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

  4. #4
    ATH0 quzah's Avatar
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    '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.
    Hope is the first step on the road to disappointment.

  5. #5
    {Jaxom,Imriel,Liam}'s Dad Kennedy's Avatar
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    Quote Originally Posted by quzah View Post
    '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???

  6. #6
    ATH0 quzah's Avatar
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    Maybe. He didn't seem to read it when you wrote it, a half an hour before his last post, so maybe if we keep saying it, he'll eventually figure it out.


    Quzah.
    Hope is the first step on the road to disappointment.

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    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.

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    Quote Originally Posted by Martin_T View Post
    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
    }

  9. #9
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    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)

  10. #10
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    Are you asking a question or giving suggestion to forum members ??

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    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.

  12. #12
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    Dude in C you cannot do that this is simple string comparison with == operator

  13. #13
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    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.

  14. #14
    C++ Witch laserlight's Avatar
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    I think the two of you are actually in agreement
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