CODEWITHSUNDEEP
c++gccc++11implicit-conversiontype-conversion
magu_
magu_

Reputation: 4876

Does gcc convert uint8_t to int for single values?

If I compile the following program:

#include <vector>
#include <cstdint>
#include <stdio.h>

int main() {
    constexpr std::size_t N = 10;
    uint8_t int8Value = 42;

    std::vector<int> IntVector(N, 0);
    for (int & ele:IntVector) {
        ele += int8Value;
    }

    std::vector<uint8_t> Int8Vector(N, 0);
    for (uint8_t & ele:Int8Vector) {
        ele += int8Value;
    }

    for (std::size_t i = 0; i < N; i++) {
        printf("%i %i\n",IntVector[i],Int8Vector[i]);
    }
}

with g++ test.cpp -o test -std=c++11 -Wconversion on gcc 4.9 it spits out the folowing warning:

test.cpp: In function ‘int main()’:
test.cpp:16:7: warning: conversion to ‘uint8_t {aka unsigned char}’ from ‘int’ may alter its value [-Wconversion]
ele += int8Value;
   ^

So if I understand it correctly this means that the compiler converts the single value of uint_8 to int? Is this because of memory alignment?

On the other hand if I try something like:

uint8_t int8Value = 342;

it throws me an overflow warning and trancates the result. Also 

printf("%i\n",sizeof(int8Value));

returns the expected 1.

Am I missing something obvious?

Upvotes: 1

Views: 4932

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409482

Yes, arithmetic expression causes conversion to int. See e.g. this reference.

Upvotes: 6

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