Using size_t variable in C99

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I read recently about the use of the size_t variable in C99.
I sow many people using it in dynamic arrays , some people for loop and others everywhere (eg in function vars).

My main question is this:

I know that size_t takes unsigned values only and that can used in replace of usigned long variable type.

I recently used it and I can't understand if I took the main idea. Can I use size_t everywhere I need a unsigned int or unsigned long type of variable?

My piece of code:

C

//in main function
FILE *input = fopen("input.txt","r");
size_t size = read_number(input);
int *array = malloc(size * sizeof *array);

//read the size of dynamic allocation 
size_t read_number(FILE *input) {
size_t size = 0;
fscanf(input, "%d",&size);

return size;
}

Posted 13-Jan-15 5:23am

GeorgeGkas

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Sergey Alexandrovich Kryukov · Accepted Answer · 2015-01-13T05:45:00

Solution 1

This is not a "variable", this is a type: http://www.cplusplus.com/reference/cstring/size_t[^].

You should use this type to calculate data sizes, not unsigned long or anything else.
~~Also, sizeof *array is a size of an element, not a size of array. To get a size of array, multiply element size by the number of elements.~~ (You actually correctly get the size of the data to allocate.)

—SA

Posted 13-Jan-15 5:45am

Sergey Alexandrovich Kryukov

Updated 13-Jan-15 6:03am

v3

Comments

[no name] 13-Jan-15 11:45am

so my code is false?

Sergey Alexandrovich Kryukov 13-Jan-15 11:47am

I would say, it's wrong. :-)
—SA

[no name] 13-Jan-15 11:50am

ok thank you. I need to learn how to use the size_t. Maybe when I'll really need it! :) For now the classic unsigned int helps!

[no name] 13-Jan-15 11:51am

Using int my code isn't stable? And replace int *array = malloc(size * sizeof *array); with int *array = malloc(size * sizeof(int)); ?

Sergey Alexandrovich Kryukov 13-Jan-15 12:06pm

Sorry, I probably overlooked something in your code. The size of allocated memory is calculated correctly, assuming the size value is correct.
—SA

Sergey Alexandrovich Kryukov 13-Jan-15 11:54am

Don't use unsigned int. Just use size_t instead.
—SA

Aescleal 13-Jan-15 11:48am

sizeof *p is the size of whatever p is pointing at.

Sergey Alexandrovich Kryukov 13-Jan-15 11:51am

But the required data size is unknown.
—SA

Aescleal 13-Jan-15 11:58am

Maybe a more direct way of saying it would be:

int *p = malloc( n * sizeof(int) );

but seeing as the expression *p has a type of int then:

int *p = malloc( n * sizeof *p );

is effectively the same expression.

Sergey Alexandrovich Kryukov 13-Jan-15 12:05pm

Right. Hm... The question wasn't edited, so I probably overlooked it in the OP's code. What I though I saw was something different.
—SA

Aescleal · Accepted Answer · 2015-01-13T05:45:00

size_t will usually be the biggest unsigned integer type on your system. If you use it wherever you're dealing with sizes of objects (duh, clue in the name!) and loop indices your code will probably not go horribly wrong on you and you might be protected from size of integer related portability problems.

However it can open cans of worms if you don't know what you're doing. Your code for example tells fprintf that you're trying to read a signed integer into a size_t - not necessarily what you wanted! Add to that on 32 bit systems a signed integer is likely to be half the size of a size_t so you've got a potential problem on your hands.

As you're using C99 (the //comments give it away) then you can get over the printf family disaster by using the (IIRC) %z format specifier which maps to whatever size_t is defined as.