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there is a question puzzles me.In the code below msg and quote are both char array.why sizeof(msg) and sizeof(quote) gave me different result?

Code: 

#include <stdio.h>

void fortune_cookie(char msg[])
{
    printf("Message reads: %s\n",msg);
    printf("msg occupies %i bytes\n",sizeof(msg));

}

int main()
{
    char quote[] = "Cookies make you fat";

    printf("quote occupies %i bytes\n",sizeof(quote));

    fortune_cookie(quote);
    return 0;
}

Result:

quote occupies 21 bytes                                                                                                              
Message reads: Cookies make you fat
msg occupies 8 bytes
timrau
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Cedared
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    `void fortune_cookie(char msg[])` is the same as `void fortune_cookie(char* msg)`, so you get sizeof(char*), your result is 8 which means you're using 64-bit machine, right? – Mine Dec 12 '13 at 06:36
  • this question has been here many times. Could you please search a bit before asking? – V-X Dec 12 '13 at 08:51

4 Answers4

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sizeof an array equals total of sizeof each element.

When you pass an array to a function, it passes the pointer.

In your case:

quote occupies 21 bytes: Means your char array has 21 elements and each element size is 1. msg occupies 8 bytes: Means your parameter is a pointer, it will hold 8 bytes on 64-bit machine and 4 bytes on 32-bit machine.

Yu Hao
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Wei Shao
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Here char quote[] = "Cookies make you fat"; memory is allocated for the message inside quote[] that's why the size of the quote[] is 21 bytes.

void fortune_cookie(char msg[]) Here char msg[] is same as char *msg and its a pointer to the quote, because you are passing quote to fortune_cookie(quote); Size of pointer is depending on the machine. I think you are using 64 bit machine that's why you are getting 8 as its size.

Chinna
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The quote is an array in the scope of the main function. The sizeof an array is the number of bytes that it occupies, which is the length of your string + 1 for the null byte at the end.

When you pass it to the fortune_cookie, it is demoted to a pointer. The size of a pointer is 4 bytes in 32bit architectures and 8 bytes in 64bit architectures. To see the same output, you can change the signature of the fortune_cookie:

void fortune_cookie(char msg[21])

But that is generally not practical.

perreal
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Hiiii..

When sizeof() is applied to the name of an array the result will be the size of the whole array in bytes. Here the name of an array is converted to a pointer to the first element of the array.

Now when we pass this array as a Parameter to the function the receiving function variable is just the pointer , which will point the same array, we can not pass an array as Call by Value.

So "sizeof(msg)" will always display Size of a pointer.

For refrence see .:http://en.wikipedia.org/wiki/Sizeof and Stack

Community
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Kalpit S.
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    -1: Why do you say a static array? Also, the call by value comment is irrelevant. – perreal Dec 12 '13 at 06:53
  • Its because dynamic array can be resized using realloc. And If i want to get the size I need to track myself about the actual size calculation. For refrence you can c [link](http://en.wikipedia.org/wiki/Sizeof) and [link](http://stackoverflow.com/questions/8717267/how-to-get-the-length-of-a-dynamically-created-array-of-structs-in-c) – Kalpit S. Dec 12 '13 at 06:58
  • I think the correct term is just array. When you say static it refers to something like `static char array[21]`. – perreal Dec 12 '13 at 07:01
  • So please edit your answer such that it becomes correct. Otherwise it might confuse others, later. – Jens Gustedt Dec 12 '13 at 07:57