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Convert char* to uint8_t [2 methods]

C++

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In this article, we have covered different approaches to convert char* to uint8_t.

Table of contents:

  1. Approach 1: Using atoi
  2. Approach 2: Copy bits

Approach 1: Using atoi

In this approach, we convert char* to integer (32 bit) using the standard function atoi() and convert the 32 bit integer to 8 bit unsigned integer (uint8_t).

atoi() is defined in stdlib.h header file.

char* input = "110";
uint8_t input2 = (uint8_t)atoi(input);

Following is the complete C++ code to convert char* to uint8_t using atoi():

#include <stdlib.h>

int main() {
  char* input = "110";
  uint8_t input2 = (uint8_t)atoi(input);
  return 0;
}

Approach 2: Copy bits

Char datatype is of size 8 bits and uint8_t is of 8 bits as well. char* is a memory address of 32 bits or 64 bits pointing to a data (string) of variable size.

If you wish to copy the first 8 bits binary representation of a char using char*, then the approach is to assign the uint8_t variable the value pointed to by char*.

char input1;
char * input2 = &input1;
uint8_t input3 = *input2;

Following is the complete C++ code following this approach:

#include <stdlib.h>

int main() {
  char input1 = 'A'; // Value 65
  char * input2 = &input1;
  uint8_t input3 = *input2;
  return 0;
}

With this article at OpenGenus, you must have the complete idea of how to convert a char* to uint8_t.

Geoffrey Ziskovin

Geoffrey Ziskovin

Geoffrey Ziskovin is an American Software Developer and Author with an experience of over 30 years. He started his career with Haskell and has interviewed over 700 candidates for Fortune 500 companies

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Convert char* to uint8_t [2 methods]
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