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

  • 25-06-2008 10:32PM
    #1
    AARRRGH
    Closed Accounts Posts: 12,382 ✭✭✭✭


    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.


Comments

  • #2
    AARRRGH
    Closed Accounts Posts: 12,382 ✭✭✭✭AARRRGH


    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?


  • #3
    fasty
    Registered Users, Registered Users 2 Posts: 981 ✭✭✭fasty


    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.


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