How come a printf is performed after a true condition was tested?

Question

I'm a beginner in C ,so please forgive me if this question is stupid or was asking weirdly. I'm reading C primer plus and one of the examples in Chapter-8 is some loop that testing whether the user entered - `a newline character or not` ,which I couldn't understand. The code is short so I will show it to you: ``` int main(void) { int ch; /* character to be printed */ int rows, cols; /* number of rows and columns */ printf("Enter a character and two integers:
"); while ((ch = getchar()) != '
') { if (scanf("%d %d",&rows, &cols) != 2) break; display(ch, rows, cols); while (getchar() != '
') continue; printf("Enter another character and two integers;
"); printf("Enter a newline to quit.
"); } printf("Bye.
"); return 0; } void display(char cr, int lines, int width) // the function to preform the printing of the arguments being passed ``` What i dont understand is right here: ``` while (getchar() != '
') continue; printf("Enter another character and two integers;
"); printf("Enter…

Accepted Answer

Because there are no braces after the `while` statement, only the very next line is included in the loop. So, `continue` continues the while loop until a new line character is found and then the execution continues to the `printf` statements.

It is equivalent to this:

```
 while (getchar() != '
')
 {
    continue;
 }
```

How come a printf is performed after a true condition was tested?

Solution

Problem