This following code:
printf("%d. %-10s:", 1, "Test");
produces this output:
1. Test :// 14 characters long
I would like the output length of the entire format `%d. %-10s:" to be exactly 10 characters like this:
1. Test: // 10 characters
Note:
The numbers vary in length, it could be 1 or 100, so I can't deduce it's length from the output.
How can I do that?
You need to use two steps:
char buffer[20];
snprintf(buffer, sizeof(buffer), "%d. %s:", 1, "Test");
printf("%-*s", 10, buffer);
The snprintf()
operation give you a string 1. Test:
in buffer
; note that it includes the :
in the output, and assumes no trailing blanks on the string "Test"
). The printf()
operation formats the string left justified ( -
) in a length of (at least) 10 (the *
in the format and the 10 in the argument list) onto standard output. Presumably, something else will appear after this output on the same line; otherwise, there's no obvious point to the blank padding.
For full information, see:
This covers the basic operation of the *printf()
family of functions (but does not list the interfaces to the v*printf()
or *wprintf()
families of functions).
The code in the question and in the answer above is all done with constants. A more realistic scenario would be:
void format_item(int number, const char *text, int width)
{
char buffer[width+1]; // C99 VLA
snprintf(buffer, sizeof(buffer), "%d. %s:", number, text);
printf("%-*s", width, buffer);
}
Note that this code truncates the formatted data if the number plus the string is too long. There are ways around that if you work a bit harder (like adding more than 1
to width
in the definition of buffer
— maybe add 15 instead of 1).
You might write:
format_item(1, "Test", 10);
or:
char *str_var = …some function call, perhaps…
int item = 56;
format_item(++item, str_var, 20);
etc.
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