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Editorial Team
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Asked: 2026-05-13T10:56:42+00:00

From the lectures notes of a course at university, on call-by-value: void fun(int *ip)

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From the lectures notes of a course at university, on “call-by-value”:

void fun(int *ip)

{

*ip =100;

}

called by 

int n=2;  

int *np;

np = &n; 

fun(np);

would change the value of n to 100.

When we say “int *ip”, what exactly do we mean? A pointer of type integer? If so, when we call fun() with np as its argument, shouldn’t there be an error as np has the address of n, which is not an integer?

And then, we change the value of ip to 100, so doesn’t that mean that n now has the value that’s in the “memory slot” with the address 100? I am sure I am missing something. 🙂

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1 Answer

  1. Editorial Team

    A pointer of type integer?

    No, a pointer to an integer.

    when we call fun() with np as its argument, shouldn’t there be an error as np has the address of n, which is not an integer?

    n is an integer so there’s no problem. &n, np and ip all have the same type in your code: int*.

    And then, we change the value of ip to 100

    No … we change the value of *ip, not of ip. That is, we change the value that ip points to (which is also sometimes called the pointee).

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