C99 standard has integer types with bytes size like int64_t. I am using Windows's %I64d
format currently (or unsigned %I64u
), like:
#include <stdio.h>
#include <stdint.h>
int64_t my_int = 999999999999999999;
printf("This is my_int: %I64d
", my_int);
and I get this compiler warning:
warning: format ‘%I64d’ expects type ‘int’, but argument 2 has type ‘int64_t’
I tried with:
printf("This is my_int: %lld
", my_int); // long long decimal
But I get the same warning. I am using this compiler:
~/dev/c$ cc -v
Using built-in specs.
Target: i686-apple-darwin10
Configured with: /var/tmp/gcc/gcc-5664~89/src/configure --disable-checking --enable-werror --prefix=/usr --mandir=/share/man --enable-languages=c,objc,c++,obj-c++ --program-transform-name=/^[cg][^.-]*$/s/$/-4.2/ --with-slibdir=/usr/lib --build=i686-apple-darwin10 --program-prefix=i686-apple-darwin10- --host=x86_64-apple-darwin10 --target=i686-apple-darwin10 --with-gxx-include-dir=/include/c++/4.2.1
Thread model: posix
gcc version 4.2.1 (Apple Inc. build 5664)
Which format should I use to print my_int variable without having a warning?
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