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iOS Objective-C Basics (Retired) Functional Programming in C Scope

return types

how did the function scope_it_out return a value called "gamma" even though the function 'scope_it_out' is declared as void?

2 Answers

William Li
William Li
Courses Plus Student 26,868 Points
void scope_it_out() {
    char bravo[] = "gamma";
    printf("%s\n", bravo);

"gamma" was printed on the output console because the printf() function is doing its job, but it's important to remember that printf() and whatever results it generated are NOT the return value, in order for a function to have a return value, the return keyword must be used explicitly.

It's also worth knowing that, when calling a function with return value by itself, nothing will be outputted on the screen unless you pass its return value as argument to a printf() function.

Thanks for the help William.It fixed my doubt.

Hi Surya,

Here is the scope_it_out function definition:

void scope_it_out() {
    char bravo[] = "gamma";
    printf( "%s\n", bravo );

Instead of returning a value, a call is made to printf, which prints the contents of the character array bravo. The scope_it_out function itself never returned a value. The call to printf is what caused the string gamma to be displayed.

Thanks Zack for clarifying my doubt.