Unable to get output..........

This is a discussion on Unable to get output.......... within the C Programming forums, part of the General Programming Boards category; Code: main() { int i=3; printf("value of i=%d",i); printf("address of i=%u",&i); printf("value of i=%d",*(&i)); }...

  1. #1
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    Unable to get output..........

    Code:
    main()
    {
     int i=3;
     printf("value of i=%d",i);
     printf("address of i=%u",&i);
     printf("value of i=%d",*(&i));
    }
    C enlightened

  2. #2
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    There are no error messages,I just cannot see the output screen.
    C enlightened

  3. #3
    Registered User Codeplug's Avatar
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    stdout is buffered and will display when it reaches a newline or is flushed.

    gg

  4. #4
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    Hello codeplug,can you explain what is stdout?Is it a pre processor directive like stdio?
    C enlightened

  5. #5
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    Thanks,got the output,just had to use a pre-processor directive.
    C enlightened

  6. #6
    Registered User Codeplug's Avatar
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    >> can you explain what is stdout?
    "stdout" is the stream that printf() prints to.

    gg

  7. #7
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    Quote Originally Posted by sameertelkar View Post
    Thanks,got the output,just had to use a pre-processor directive.
    The only preprocessor directive needed is #include <stdio.h>

    If you do this you will get an output:

    Code:
    #include <stdio.h>
    
    
    main()
    {
         int i=3;
         printf("value of i=%d",i);
         printf("address of i=%u",&i);
         printf("value of i=%d\n",*(&i));
    return;
    }
    
    

  8. #8
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    As long as we are making improvements to the code, we may as well fix everything:
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int main(void)
        {
        int i = 3;
    
        printf("value of i = %d\n", i);
        printf("address of i = %p\n", (void*)&i);
        printf("value at %p = %d\n", (void*)&i, *(&i));
    
        return EXIT_SUCCESS;
        }

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