CODEWITHSUNDEEP
cconstants
Fakher Mokadem
Fakher Mokadem

Reputation: 1099

Declaring function arguments as const in C

What is the use of declaring arguments of a function in C as const? Does it have an impact on performance?

e.g. why would you prefer f1 to f2? Or f2 to f1?

void f1(const char* arg) {
/* some code */
}

void f2(char* arg) {
/* some code */
}

Upvotes: 1

Views: 1900

Answers (4)

kiran Biradar
kiran Biradar

Reputation: 12732

You prefer 

void f1(const char* arg)

If you don't want to modify the object pointed by arg through arg. But you can modify the arg itself.

You prefer 

void f2(char* arg)

If you want to modify the object pointed by arg through arg.

Upvotes: 1

babon
babon

Reputation: 3774

const char* arg means that arg is a pointer to a constant char. If you try to change the data that arg points to, you will get an error saying something like: error: assignment of read-only location ‘*arg’. However, you can change the address contained in arg.

Whereas, by dropping the const keyword, you state that the data which arg points to can be changed.

Your preference should be based on your needs. Have a look at this for more information.

In response to your comment: char * const arg means means that arg is a constant (pause a second) pointer to a char. So, you can't change the address contained in arg and you can change the data residing at that address.

Upvotes: 2

chqrlie
chqrlie

Reputation: 144695

in void f1(const char *arg) the argument arg itself is not const qualified, it is defined as a pointer to an array of char that the function should not modify. This is very useful for the reader to understand at first glance that the function f1 does not modify the string it receives.

For example strcpy is declared in <string.h> as:

char *strcpy(char *dest, const char *src);

The array pointed to by the first argument is modified, it is the destination array, the second argument points to the source array, which is not modified.

This type of const qualification is viral: once you define an argument as pointing to a const object, you can only pass it to functions that have similarly const qualified arguments.

Regarding the impact on performance, there is no downside. Some compilers might even benefit from this qualifications and generate better code.

So in your example, if f1 does not modify the array pointed to by arg, unless you need to store f1 in a function pointer of type void (*)(char*) and cannot change this type, I strongly advise to use:

void f1(const char *arg);

Upvotes: 3

0___________
0___________

Reputation: 67476

void foo(const char *arg)

declares the pointer to the const char. You cant change the referenced object, but you can change the pointer

void foo(char * const arg)

declares the const pointer to the char. You cant change the the pointer, but you can change the referenced object

void foo(const char * const arg)

declares the const pointer to the const char. You cant change the the pointer and the referenced object

declaring parameters allows many compile time optimizations and may have a significant impact on the performance of the generated code.

It is considered a very good practice to use const for the parameters and variables which should not change. Is is called the "const correctness"

Futher reading: restrict keyword.

Upvotes: 2

Related Questions