How to print a guint64 value when using glib?

Question

Problem I am making use of the GLib 2.0 library, and declared a gunit64 variable. I wish to print its value to screen, but its not working properly. Code Consider the following code snippet as an example. I declare a guint64 variable, and try to print its value. ``` guint64 myValue = 24324823479324; printf("My Value: %d
", myValue); ``` Warning I get this warning: ``` warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘guint64’ ``` Output I get a strange negative number on screen: My Value: -1871285220 Further Comments I tried to search the API's documentation, and I found the following under guint64: An unsigned integer guaranteed to be 64 bits on all platforms. Values of this type can range from 0 to G_MAXUINT64 (= 18,446,744,073,709,551,615). To print or scan values of this type, use G_GINT64_MODIFIER and/or G_GUINT64_FORMAT. Therefore, I assume I have to use either the modifier or format definitions. However, the documentation does not show how to use th…

Accepted Answer

You've found the proper macro for this. Here's how to use it:

```
printf("My value: %" G_GUINT64_FORMAT "
", myValue);
```

Note that the macro is a quoted string literal, so the above is the proper syntax. Also note that the `%` is not part of the macro.

Your number appears negative since you're using `%d`, which expects an `int`, and the bits of your number, when viewed as a smaller, signed `int`, encode something negative.

How to print a guint64 value when using glib?

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