C passing variable by reference

Question

I have the following code:

main()
{
 uint8_t readCount;
 readCount=0;
 countinfunc(&readCount);
}

countinfunc(uint8_t *readCount)
{
 uint8_t i;
 i = readCount;
 ....
}

Problem is that when it enters in the function the variable i has a different value then 0 after the assignment.

`warning: assignment makes integer from pointer without a cast [enabled by default]` and `warning: return type defaults to int` — Karoly Horvath, Oct 25 '12 at 13:15

score 9 · Accepted Answer · answered Oct 25 '12 at 13:11

It's because in countinfunc the variable is a pointer. You have to use the pointer dereference operator to access it in the function:

i = *readCount;

The only reason to pass a variable as a reference to a function is if it's either some big data that may be expensive to copy, or when you want to set it's value inside the function to it keeps the value when leaving the function.

If you want to set the value, you use the dereferencing operator again:

*readCount = someValue;

score 3 · Answer 2 · answered Oct 25 '12 at 13:10

3

countinfunc(uint8_t *readCount)
{
 uint8_t i;
 i = *readCount;
 ....
}

answered Oct 25 '12 at 13:10

MOHAMED

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score 3 · Answer 3 · answered Oct 25 '12 at 13:10

3

replace

i = readCount;

with

i = *readCount;

answered Oct 25 '12 at 13:10

Bhavik Shah

5,125
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score 2 · Answer 4 · edited Mar 05 '13 at 22:36

2

You're setting i to the address of readCount. Change the assignment to:

i = *readCount;

and you'll be fine.

edited Mar 05 '13 at 22:36

answered Oct 25 '12 at 13:11

Anders R. Bystrup

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I can't believe I didn't try that. – dare2k Oct 25 '12 at 13:15

score 1 · Answer 5 · edited May 23 '17 at 12:19

1

just replace

i=readCount by i=*readCount

You cannot assign (uint8_t *) to uint8_t

For more on this here is the link below Passing by reference in C

edited May 23 '17 at 12:19

Community

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1

answered Oct 25 '12 at 13:16

Omkant

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C passing variable by reference

5 Answers5

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