Sizeof array increases by one byte after passing it to a function

AARRRGH · 25-06-2008 9:32pm #1

Hello

I was wondering if any of you have an explanation for this.

int main()
{
	unsigned char a[] = {0x00, 0x80, 0x81};
	unsigned char b[] = {0x00, 0x18, 0x00};

	printf("sizeof a: %d\n", sizeof a);

	multiply(a, b, f);
}

This outputs the text "sizeof a: 3".

The multiply function is then called:

void multiply(unsigned char *a, unsigned char *b) {

	printf("sizeof a: %d\n", sizeof a);
}

This outputs the text "sizeof a: 4".

???

Can any of you think of an explanation for this?

Thanks in advance.

AARRRGH · 25-06-2008 9:48pm

Thinking about this some more, I can see it's the size of the pointer to the array I'm seeing in the multiply function.

Does anyone know how I can get it to output the size of the array?

fasty · 25-06-2008 9:50pm

The first sizeof is getting the size of an array, which is 3, the sizeof in multiply is getting the sizeof a pointer to an unsigned char.

EDIT, Oh you realised... There isn't really a way to tell the length in C or C++, You'd have to pass size values to the multiply function.

Sizeof array increases by one byte after passing it to a function

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