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Type: Posts; User: seten

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  1. I think I understand what youre saying on its own...

    I think I understand what youre saying on its own but I must not because I don't see how that contradicts my statement.
    Foo returns (q+y) which in this case is (x+y) or (1), and I understand that...
  2. So is q automatically 0 since we never assigned...

    So is q automatically 0 since we never assigned it a value? and is foo(x) the just the value that foo returns if it used x as the value instead of q?

    So when it jumps to foo which prints q=0 and...
  3. This should be an easy one for yall (beginner)

    Okay, help me understand whats going on here.



    #include <stdio.h>

    int foo (int q) { /* so here we're declaring a function called foo which returns an integer.
    (int q) means that our foo...
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