How to jump out of a C++ code block?

Question

This is what I would like to do:

{
    ...
    if(condition)
        break;
    ...
}

This works for a loop. I would like something similar

Answer 1

In C++11 the best way to achieve this is use a anonymous lambda function and replacing break with return

[&](){
    ...
    if(condition)
        return;
    ...
 }();

Note that [&] captures all variables by reference, if you don't use any just replace it with []

Answer 2

I'm afraid I'm unleashing the wrath of a coding god against me right now, but this nice nasty macro could help people do this easier than before (based on @jleahy's answer):

//example use:
// { __breakable__
//    ...
//    if (condition)
//        break;
//    ...
//    if (condition2)
//        break;
//    ...
// }
//WARNING: use only with a standalone block. To make if, try or catch block
//         breakable use double braces
//         if (cond) {{ __breakable__ ... }}
//         try {{ __breakable__ ... }} catch (...) {{ __breakable__ ... }}
#define __breakable__ }switch(0){default:

Answer 3

"I would like something similar for a simple block of code."

Use return after a condition is met, and return control back to the caller.

void MyWorkerClass::doWork(Item & workItem) {
    // ...
    if(noMoreWorkToDo)
        return;
    // ...
}

Answer 4

Here just some additional possibilities:

for(..)
{
    continue;//next loop iteration
}

void mymethod()
{
    ...
    return;
    ...
}

Probably you should create sub-methods for the problematic block of code were you wanted to use goto and leave the block of code by the usage of return.

Answer 5

This one:

{
    // ...

    if (!condition)
    {
        // ...
    }
}

This will avoid goto to jump out of a block of code.

Answer 6

Here's one way:

switch(0) {
default:
    /* code */
    if (cond) break;
    /* code */
}

(please never do this)