1 byte integer doesn't convert i/o formats

Question

I wrote the code below which inputs a number in hex format and outputs it in decimal form:-

#include<iostream>
#include<iomanip>
#include<stdint.h>

using namespace std;

int main()
{
  uint8_t c;
  cin>>hex>>c;
  cout<<dec<<c;
  //cout<<sizeof(c);
  return 0;
}

But when I input c(hex for 12), the output was again c(and not 12). Can somebody explain?

score 6 · Accepted Answer · answered Dec 09 '12 at 14:16

6

This is because uint8_t is usually a typedef for unsigned char. So it's actually reading 'c' as ASCII 0x63.

Use int instead.

#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
    int c;
    cin>>hex>>c;
    cout<<dec<<c<<'\n';
    return 0;
}

Program output:

$ g++ test.cpp
$ ./a.out
c
12

answered Dec 09 '12 at 14:16

Dietrich Epp

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@dietrich...so is there a way in which i can input a 1 byte data into in hex format into an 1 byte integer and it behaves as integer and not char – AvinashK Dec 09 '12 at 20:11
Go ahead and click "Ask Question" at the top. – Dietrich Epp Dec 09 '12 at 20:29

score 4 · Answer 2 · answered Dec 09 '12 at 14:17

4

This is an unfortunate side effect of the fact that uint8_t is actually unsigned char. So when you store c, its storing the ASCII value of c (99 decimal), not the numeric value 12.

answered Dec 09 '12 at 14:17

Doug T.

64,223
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score 0 · Answer 3 · answered Mar 05 '21 at 04:32

uint8_t is an alias of unsigned char and unfortunately ostream tries to output it as a character. This has been fixed in C++20 std::format:

#include <format>
#include <iostream>
#include <stdint.h>

int main() {
  uint8_t n = 42;
  std::cout << std::format("{}", n);
}

Output:

1 byte integer doesn't convert i/o formats

3 Answers3