Array as parameter doesn't maintain size

Asked Oct 23 '16 at 22:14

Active Oct 23 '16 at 22:26

Viewed 35 times

I'm testing sizeof macro in C/C++.

My main code is as follows

 void ShowSize(int RA[]);

 main()
 {
     int MyRA[10] = {1,2,3,4,5,6,7,8,9,0};

     printf("size of MyRA is %d\n", sizeof(MyRA) );

     ShowSize(MyRA);
 }

 void ShowSize(int RA[])
 {
     printf("Size in here is different %d\n", sizeof(RA));
 }

I would have thought the size is the same in both cases but it is not...

How can I make both printf statements show the same value? I've tried dereferencing the value inside the function since it's a pointer, right, but my values are not consistent.

edited Oct 23 '16 at 22:26

asked Oct 23 '16 at 22:14

fifamaniac04

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1

It doesn't yes. Good observation. – πάντα ῥεῖ Oct 23 '16 at 22:15
2

"I'm testing `sizeof` macro"? `sizeof` is not a macro. – AnT stands with Russia Oct 23 '16 at 22:16
1

Make a second parameter to pass the length. And don't call your language "C/C++". – deviantfan Oct 23 '16 at 22:16
1

You can only make it work by coincidence. For example, on a 32-bit platform, `int MyRA[1] = { 1 };` will print 4 twice; on a 64-bit platform, `int MyRA[2] = { 1, 2 };` will print 8 twice. But that's coincidence. (More generally: `int MyRA[sizeof(void *)/sizeof(MyRA[0])];` will print the same value twice. – Jonathan Leffler Oct 23 '16 at 22:16
@AnT would you just call it a built-in function? – fifamaniac04 Oct 23 '16 at 22:17
4

@fifamaniac04: In C and C++ `sizeof` is an *unary operator*. – AnT stands with Russia Oct 23 '16 at 22:18
`sizeof` is an operator! It cannot be a macro for obvious reasons! – too honest for this site Oct 23 '16 at 22:19
Similar previous question [here](http://stackoverflow.com/questions/8269048/length-of-array-in-function-argument) – Weather Vane Oct 23 '16 at 22:19
Note that your code is archaic C code and not C++ code — because you've not specified the return type of `main()`. That makes it C89/C90 code at the latest; C99 and C11 both require the return type, and C++ 'always' has required it too. The missing return type on `ShowSize()` is dubious too — does your code compile even with a C compiler? – Jonathan Leffler Oct 23 '16 at 22:20
Your `printf` conversion type specifier is wrong, your code invokes _underfined behaviour_. Read the documentation of functions you use. – too honest for this site Oct 23 '16 at 22:22
lol, yes my code compiles and runs... – fifamaniac04 Oct 23 '16 at 22:25
@Olaf there is nothing wrong with my printf statement – fifamaniac04 Oct 23 '16 at 22:27
`printf("size of MyRA is %zu\n", sizeof MyRA);`. The compiler is not obliged to check the format specifiers against the arguments passed, although some do. – Weather Vane Oct 23 '16 at 22:28
@fifamaniac04: You want to first read the [standard](http://port70.net/~nsz/c/c11/n1570.html#7.21.6.1p7) about this and then come back! – too honest for this site Oct 23 '16 at 22:31
C does not have parameters of array type. [comp.lang.c FAQ](http://www.c-faq.com) section 6. – Keith Thompson Oct 23 '16 at 22:33

Array as parameter doesn't maintain size

0 Answers0