How can you prematurely exit from a function without returning a value if it is a void function? I have a void method that needs to not execute its code if a certain condition is true. I really don't want to have to change the method to actually return a value.
This question is related to
c++
Use a return statement!
return;
or
if (condition) return;
You don't need to (and can't) specify any values, if your method returns void.
You mean like this?
void foo ( int i ) {
if ( i < 0 ) return; // do nothing
// do something
}
void foo() {
/* do some stuff */
if (!condition) {
return;
}
}
You can just use the return keyword just like you would in any other function.
You mean like this?
void foo ( int i ) {
if ( i < 0 ) return; // do nothing
// do something
}
void foo() {
/* do some stuff */
if (!condition) {
return;
}
}
You can just use the return keyword just like you would in any other function.
void foo() {
/* do some stuff */
if (!condition) {
return;
}
}
You can just use the return keyword just like you would in any other function.
You mean like this?
void foo ( int i ) {
if ( i < 0 ) return; // do nothing
// do something
}
You mean like this?
void foo ( int i ) {
if ( i < 0 ) return; // do nothing
// do something
}
void foo() {
/* do some stuff */
if (!condition) {
return;
}
}
You can just use the return keyword just like you would in any other function.
Source: Stackoverflow.com