Can't make printf happy. Rejecting format strings of "i" and "l"

Question

long int triagedOut = 0;

printf("%i", triagedOut);

main.c|89|warning: format '%i' expects argument of type 'int', but argument 2 has type 'long int' [-Wformat=]|

I actually get that message twice for the same line. ???

OK, so I will change it to the long int specifier "l" and, then...

printf("%l", triagedOut);

main.c|89|warning: conversion lacks type at end of format [-Wformat=]|

main.c|89|warning: too many arguments for format [-Wformat-extra-args]|

And I get these two messages twice, too.

I don't even know how there could be too many of anything. Just 1 of everything. What's it trying to tell me?

If you want to print a long int number you should use `%li` to print it. [C types](https://en.wikipedia.org/wiki/C_data_types) — dinhanhx, Feb 27 '21 at 07:20
"*how there could be too many of anything*" What the compiler is telling you is that `"%l"` is not recognized as a valid format specifier, and without any such specifier the `triagedOut` argument passed into the call is one too many. Same warning as for `printf("hello world", triagedOut);` for example. — dxiv, Feb 27 '21 at 07:58

score 4 · Answer 1 · edited Feb 27 '21 at 08:57

You need to use %ld for long int.

printf("%ld", triagedOut);

Note also these remarks:

Format %i indeed expects an int argument, and you are passing a long int, which is a different type and may have a different size, depending on the target platform.
Format %l is incomplete: l is just a length modifier, you also need a conversion specifier such as i, d, u, x... hence the proposed correction %ld. i and d are equivalent for printf(), but they have a different meaning for scanf(): %i converts an integer expressed in decimal, octal or hexadecimal as determined by its initial characters, whereas %d only converts an integer expressed in decimal.
```
 int i;
 sscanf("0x10", "%i", &i);  // sets i to 16
 sscanf("0x10", "%d", &i);  // sets i to 0
 sscanf("010", "%d", &i);   // sets i to 10
 sscanf("010", "%i", &i);   // sets i to 8
```
The reason you get the messages twice might be that these are warning messages, not preventing the compilation from proceeding, and your IDE might compile 2 different versions of the executable, one in debug mode and one in production mode.

score 2 · Answer 2 · answered Feb 27 '21 at 07:48

2

I think you should use %ld or %li instead of %l or %i for printing long integer.

long int triagedOut = 0;

printf("%li", triagedOut);

answered Feb 27 '21 at 07:48

2 Answers2