1.3-consistency-and-idioms

Chapter 1: Style

1.3 Consistency and Idioms

Exercise 1-7

Rewrite these C/C++ excerpts more clearly:

if (istty(stdin)) ;
else if (istty(stdout)) ;
    else if (istty(stderr)) ;
        else return(0);

if (retval != SUCCESS)
{
    return (retval);
}
/* All went well! */
return SUCCESS;

for (k = 0; k++ < 5; x += dx)
    scanf("%lf", &dx);

Answer:

if (istty(stdin)) {}
else if (istty(stdout)) {}
else if (istty(stderr)) {}
else
    return (0);

Here, I just added parantheses around the empty bodies of the if conditions and corrected the indentation. This should be enough to easy readability.

return retval;

Here, the return value is always retval, so there is no sense in checking it, since no other action is executed anyway.

for (k = 0; k < 5; k++)
    scanf("%lf", &dx);
    x += dx;

Here, I moved the incrementation of k to the last part of the for loop definition, as this is the idiomatic choice. Moreove, I moved x += dx into the body of the for loop, since this way is even clearer where dx is coming from and what happens with x. Also, now the for loop looks like the standart for that a seasoned programmer would recognise at a glance.

Exercise 1-8

Identify the errors in this Java fragment and repair it by rewriting with an idiomatic loop.

int count = 0;
while (count < total) {
    count++;
    if (this.getName(count) == nametable.userName()) {
        return (true);
    }
}

Answer:

for (int count = 0; count < total; count++) {
    if (this.getName(count) == nametable.userName()) {
        return (true);
    }
}

Not much to explain here. Making the loop idiomatic makes the code much easier to read and understand.