How to print hex from uint32_t? - c++

Question

The code I have been working on requires that I print a variable of type uint32_t in hexadecimal, with a padding of 0s and minimum length 8. The code I have been using to do this so far is:
printf("%08lx\n",read_word(address));
Where read_word returns type uint32_t. I have used jx, llx, etc. formats to no avail, is there a correct format that can be used?
EDIT:
I have found the problem lies in what I am passing. The function read_word is returns a value from a uint32_t vector. It seems that this is the problem that is causing problems with out putting hex. Is this a passing by reference/value issue and what is the fix?
read_word function:
uint32_t memory::read_word (uint32_t address) {
if(address>(maxWord)){
return 0;
}
return mem[address];
}
mem deceleration:
std::vector<uint32_t> mem=decltype(mem)(1024,0);

Answer 1

To do this in C++ you need to abuse both the fill and the width manipulators:
#include <iostream>
#include <iomanip>
#include <cstdint>
int main()
{
uint32_t myInt = 123456;
std::cout << std::setfill('0') << std::setw(8) << std::hex << myInt << '\n';
}
Output
0001e240
For C it gets a little more obtuse. You use inttypes.h
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main()
{
uint32_t myInt = 123456;
printf("%08" PRIx32 "\n", myInt);
return 0;
}
Output
0001e240
Note that in C, the constants from inttypes.h are used with the language string-concatenation feature to form the required format specifier. You only provide the zero-fill and minimum length as a preamble.

Answer 2

%jx + typecast
I think this is correct:
#include <stdio.h>
#include <stdint.h>
int main(void) {
uint32_t i = 0x123456;
printf("%08jx\n", (uintmax_t)i);
return 0;
}
compile and run:
gcc -Wall -Wextra -pedantic -std=c99 main.c
./a.out
Output:
00123456
Let me know if you can produce a failing test case.
Tested in Ubuntu 16.04, GCC 6.4.0.