Chris Harding
Chris Harding

Reputation: 37

Why does program not execute final printf statement?

I cannot figure out why program control does not reach the third printf, right after the for loop.

Why won't the third printf print?

If I change the for loop to while loop, it still will not print.

Here is the program and output:

main()
{
    double nc;

    printf ("Why does this work, nc = %f\n", nc);
    for (nc = 0; getchar() != EOF; ++nc)
    {
        printf ("%.0f\n", nc);
    }
    printf ("Why does this work, nc = %f", nc);
}

The output is:

Why does this work, nc = 0.000000
test
0
1
2
3
4

Upvotes: 0

Views: 2349

Answers (2)

Levon
Levon

Reputation: 143022

It works fine for me, how are you trying to termintate the program? The for-loop should end once EOF is detected as input by getchar().

EOF is Control-Z (^Z) under Windows and Control-D (^D) under Linux/Unix. Once I enter this, the loop terminates and I get the final printf() to display its output.

As a final note (as mentioned by @DanielFisher too), add a '\n' at the end of your final printf() call as it may be required by your particular implementation or otherwise the program's behavior might be undefined (thanks to @KeithThompson and @AndreyT pointing this out in the comments):

 printf ("Why does this work, nc = %f\n", nc);

Upvotes: 4

Simon
Simon

Reputation: 890

printf is buffered, that's why the final line may not be displayed. That means a call to printf may not result in a direct output as the function accumulates data before putting it in the output (your terminal).

A call to fflush after your last printf will put everything that remains in the buffer in your terminal. Also, the buffer is flushed every time you ask for a newline.

Upvotes: 0

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