C - Passing a pointer as a pointer to another function

Question

So, I have one function like the following:

void myfunction1(int *number)
{
    ...
}

And I have:

void myfunction2(int *number)
{
    ...
    myfunction1(&number);
}

When I run the code I get the error:

warning: passing argument 1 of ‘myfunction1’ from incompatible pointer type

So I changed my second function to:

void myfunction2(int *number)
{
    ...
    myfunction1(&(*number));
}

And I got:

dev(887) malloc: *** error for object 0x7fd400403810: pointer being freed was not allocated
*** set a breakpoint in malloc_error_break to debug
Abort trap: 6

Any ideas?

just myfunction1(number);

BLUEPIXY
– BLUEPIXY

2013-07-28 20:49:42 +00:00
Commented Jul 28, 2013 at 20:49 — BLUEPIXY
– BLUEPIXY, Commented Jul 28, 2013 at 20:49

Lara Bailey · Accepted Answer · 2013-07-28 20:50:56Z

2

number already has type int* so you can pass it directly to myfunction1 with myfunction1(number). The malloc error has nothing to do with any code that you have shown.

answered Jul 28, 2013 at 20:50

Lara Bailey

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Comments

hetepeperfan · Accepted Answer · 2013-07-28 20:56:00Z

2

Number is already a pointer to int. Therefore if you do &number where number is an int* you are passing a pointer to a pointer to int.

void myfunction2(int *number)
{
    myfunction1(number);
}

will do the trick

edited Jul 28, 2013 at 20:56

answered Jul 28, 2013 at 20:50

hetepeperfan

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Comments

lithiumhead · Accepted Answer · 2013-07-28 20:52:11Z

0

basically what the others are saying is this is what you should do:

void myfunction2(int *number)
{
    ...
    myfunction1(number);
}

inside myfunction2, number is already a pointer, so pass that as it is to myfunction1

answered Jul 28, 2013 at 20:52

lithiumhead

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