Why do the exec()-family functions actually return a value?

Question

In a book about Linux and Unix programming, I´ve found this statement (emphasize mine):

"All functions (exec()-family) return -1 in the case of an error. Otherwise at successful execution there is no return back to the calling program. Thus, it is redundant to check the return value; you can directly continue with the error routine."

This seems to fit with the according Linux man page:

"RETURN VALUE - The exec() functions return only if an error has occurred. The return value is -1, and errno is set to indicate the error."

Thus, If the current process image was successfully replaced with a new process image, there should be no return. Only if there was happen an error at the creation of the new image, one of the exec()-functions actually returns (-1).

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Means, in turn, instead of, f.e.:

if ( execl("/bin/foo", "foo", "bar", (char *) NULL) == -1 )
{
    perror("Execution of foo failed");
    // further error routine.
}

One can simply do:

execl("/bin/foo", "foo", "bar", (char *) NULL);

perror("Execution of foo failed");
// further error routine.

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• But why do the functions of the exec()-family have a return value then in general, if the further program execution of the caller already indicates that an error has happen?

• Wouldn´t a simple return; statement at the end of the functions and a respective return type of void be sufficient?

• Is it just to fit the common method/practice to always checking return values or has it a different purpose?

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Related:

• Please explain the exec() function and its family

• why the exec() family of functions doesn't execute the code after exec()?

• Working of exec family functions