need help please due in 3 hourse

This is a discussion on need help please due in 3 hourse within the C Programming forums, part of the General Programming Boards category; hi i want to know why is this not printf out the value Code: #include<stdio.h> #include<conio.h> #include<stdlib.h> struct computer_part { ...

  1. #1
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    need help please due in 3 hourse

    hi i want to know why is this not printf out the value

    Code:
    #include<stdio.h>
    #include<conio.h>
    #include<stdlib.h>
    
    struct computer_part
    {    
        char name;
        int quantity;
        int price;     
           
    };
    main()
    {
         struct computer_part cp[5];
         int i;
         
         printf("enter price");
         scanf("%i",&cp[i].price);
          
          printf("%i",cp[i].price);
          getch();
    
    
    }

  2. #2
    a_capitalist_story
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    Code:
    int i;
         
         printf("enter price");
         scanf("%i",&cp[i].price);
    Any guesses what the value of i is?

  3. #3
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    the value of i is the couting for array

  4. #4
    l'Anziano DavidP's Avatar
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    the value of i is the couting for array
    That might be true if you had a loop and counting going on...I do not see either
    My Website

    "Circular logic is good because it is."

  5. #5
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    No - literally - what is the value of i? If I printf'd it, what would I see?

  6. #6
    and the Hat of Guessing tabstop's Avatar
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    Nobody knows (quite literally) what the value of i is.

  7. #7
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    Code:
    #include<stdio.h>
    #include<conio.h>
    #include<stdlib.h>
    
    struct computer_part
    {    
        char name;
        int quantity;
        int price;     
           
    };
    int main()
    {
        struct computer_part cp[5];
        int i;
    
        for(i = 0; i < 5; i++) {     
          printf("enter price");
          scanf("%d",&cp[i].price);
          
          printf("\n %d",cp[i].price);
        }
        getch();
        return 0;
    }

    I wouldn't use i without it first being given a value, and it's just confusing using %i as the format specifier, and i as the index to your cp array, right on the same line of code.

    Maybe the above is what you wanted?

  8. #8
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    Quote Originally Posted by Adak View Post
    I wouldn't use i without it first being given a value,
    Right. In C, local variables are not initialised when you define ("int i") them. You need to assign it a value. Adak's example is probably what you were after.

    and it's just confusing using %i as the format specifier, and i as the index to your cp array, right on the same line of code.
    Glad I'm not the only one who didn't like that :-)

  9. #9
    Registered User claudiu's Avatar
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    Quote Originally Posted by Adak View Post
    Code:
    #include<stdio.h>
    #include<conio.h>
    #include<stdlib.h>
    
    struct computer_part
    {    
        char name;
        int quantity;
        int price;     
           
    };
    int main()
    {
        struct computer_part cp[5];
        int i;
    
        for(i = 0; i < 5; i++) {     
          printf("enter price");
          scanf("%d",&cp[i].price);
          
          printf("\n %d",cp[i].price);
        }
        getch();
        return 0;
    }

    I wouldn't use i without it first being given a value, and it's just confusing using %i as the format specifier, and i as the index to your cp array, right on the same line of code.

    Maybe the above is what you wanted?

    Damnit Adak, you ruined the whole learning experience with your genius brilliance. lol


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