CODEWITHSUNDEEP
carrays
sipsorcery
sipsorcery

Reputation: 30699

Passing C array as char* function parameter

I've got some code I'm mantaining with the following variable declaration:

char tmpry[40];

It's being used with this function:

char *SomeFunction(char *tmpryP) {
   // Do stuff.
}

The function call is:

SomeFunction(&tmpry[0]);

I'm pretty damn sure that's just the same as:

SomeFunction(tmpry);

I've even checked that the char* pointer in SomeFunction ends up pointing to the same memory location as the array in both cases.

My question is a sanity check as to whether the two function calls are identical (and therefore the original programmer was just being nasty)?

Upvotes: 7

Views: 24386

Answers (8)

mctylr
mctylr

Reputation: 5169

The C Programming FAQ section on arrays and pointers tackles this (and many other common C questions and confusions.

Upvotes: 5

Jiang Bian
Jiang Bian

Reputation: 2053

The declaration

int a[10]; int *pa;

There is one difference between an array and a pointer that must be kept in mind. A pointer is a variable, so pa=a and pa++ are legal. But an array name is not a variable; construction like a=pa and a++ are illegal

As format parameters in a function definition, char s[] and char *s are equivalent;

From: The C Programming Language 2th, Page 99-100

Upvotes: 1

Alex B
Alex B

Reputation: 84822

It may be significant, if SomeFunction is a macro and takes "sizeof" of one of its arguments, because sizeof(tmpry) may not necessarily be equal to sizeof(&tmpry[0]).

Otherwise, as others pointed out, they are exactly the same.

Upvotes: 7

Jonathan Leffler
Jonathan Leffler

Reputation: 753725

As everyone else said, the two notations are equivalent.

I would normally use the simpler one, unless there are multiple calls like this:

SomeFunction(&tmparry[0]);
SomeFunction(&tmparry[10]);
SomeFunction(&tmparry[NNN]);

Where ideally all the constants (magic numbers) would be enum (or #define) constants.

Upvotes: 3

Willi Ballenthin
Willi Ballenthin

Reputation: 6604

they are exactly the same.

someArray[i]

means exactly

*(someArray + i)

where someArray in the second case is a pointer to the memory location. Similarly, 

&someArray[i]

means exactly

(someArray + i)

In both these cases the terms are pointers to memory locations.

Upvotes: 12

sharptooth
sharptooth

Reputation: 170499

These two variants are equivalent and passing like this

SomeFunction(tmpry);

looks cleaner.

Upvotes: 1

Naveen
Naveen

Reputation: 73443

Both are one and the same although second one looks nice and clarifies the intention of passing the array to the function. But how does the SomeFunction know the size of the array being passed, is it always assumed as 40 ? I feel it is better to pass the size also as the parameter to SomeFunction.

Upvotes: 1

ojblass
ojblass

Reputation: 21620

The two are equivalent and I think SomeFunction(tmpry); is more readable.

Upvotes: 12

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