Unknown type name uint64_t and uint16_t uint8_t in Linux

Question

I'm trying to compile a program in Ubuntu environment but I'm having some error saying unknown type name 'uint64_t', unknown type name 'uint16_t' even though I have included cstdint and any other requirements I believe. It seems like the C++ library support issue.

Does anyone know how to fix this?

I'm guessing you need to use `std::uint64_t`, but can't be certain without the code. Is the code `using namespace std;` ? - Actually, just read this answer: https://stackoverflow.com/a/13643019/1730895 — Kingsley, May 22 '19 at 02:36
Please add your code, and the command used for compiling along with the environment details. — Anirudh, May 22 '19 at 03:09
you don't post the full code. You need to create a [mcve] which demonstrates your problem — phuclv, May 22 '19 at 04:19
What compiler are you using? Is it the native compiler from `apt-get` or `yum`; or is it something else?? Does it need a `-sysroot` , and is it set? — jww, May 22 '19 at 05:42

Kingsley · Answer 1 · 2019-05-22T03:20:16.067

Without seeing your code it's very hard to answer. I guess the code does not include cstdint or stdint.h, and/or it is not using the std::uint64_t syntax.

So my answer can only be a simple test/example you can run.

Compile it with:

g++ -Wall -g --std=c++11 int64_test.cpp -o int64_test -lstdc++

The Code "int64_test.cpp":

#include <cstdint>
#include <iostream>

int main( int argc, char **argv )
{
    std::uint64_t u64 = 3;
    std::int32_t  i32 = 141;

    std::cout << "u64 = " << u64 << std::endl;
    std::cout << "i32 = " << i32 << std::endl;

    return 0;
}

This code & compilation works fine on Ubuntu 18.04. I expect it will also work on Ubuntu 14.x

For the sake of completeness, a C version (since this is(was) tagged into the question too)

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>  // Must have this!

int main( int argc, char **argv )
{
    uint64_t u64 = 3;
    int32_t  i32 = 141;

    printf( "u64 = %lu\n", u64 );
    printf( "i32 = %d\n", i32 );

    return 0;
}

Davislor · Answer 2 · 2023-05-08T22:34:46.037

If you include <stdint.h>, the names should be declared in the global namespace.

If you include <cstdint>, an implementation will declare them in the std:: namespace, as std::uint8_t, etc. It is allowed, but not required, to also declare them without the std:: prefix. (Pedantically, an implementation for some architecture with a weird word size would be allowed to have no int64_t, and only declare int_least54_t and int_fast64_t, but none that I know of actually do.)

I usually #include <stdint.h>. However, a more fashionable way to do it might be:

#include <cstdint>

using std::uint8_t, std::uint16_t, std::uint64_t;

Unknown type name uint64_t and uint16_t uint8_t in Linux

2 Answers2