Why is this code printing "123" instead just 3?

Question

I've got some code here:

int n = 0;
for(n = 1;n<4;n++)
printf("%d",n);
return 0;

Why does it return '123' instead of just '3'?

I tried to Google for this issue, but I couldn't find anything useful.

Answer 1

After asking my question, I think I see what you are getting at.

What you have with

    int n = 0;
    for(n = 1;n<4;n++)
        printf("%d",n);
    return 0;

is functionally the same as

    int n = 0;
    for(n = 1;n<4;n++)
    {
        printf("%d",n);
    }
    return 0;

Since the for loop expects a statement, either a block of statements enclosed in braces, or a single one terminated with a semicolon as you have in your example. If you wanted it to just print 3 and for whatever reason wanted to use a loop just in increment a number, you would want to provide it with an empty statement as such:

    int n = 0;
    for(n = 1;n<3;n++);
    printf("%d",n);
    return 0;

or

    int n = 0;
    for(n = 1;n<3;n++){}
    printf("%d",n);
    return 0;

Both of which will only print 3.

Please note that because the variable n gets incremented and then checked, using your original bounds n < 4, the loop would end when n = 4 and thus 4 would be printed. I changed this in my last two examples. Also note the incorrect use of the term return, as some comments pointed out.