how can I convert character to integer number

Question

How could I change an array of character (string) to an integer number without using any ready function (ex atoi();) for example :-

 char a[5]=\'4534\';


        

Answer 1

You can go like this

It's working

#include <stdio.h>
#include <string.h>
#include <math.h>

int main()
{

  char a[5] = "4534";

  int converted = 0;

  int arraysize = strlen(a);

  int i = 0;
  for (i = 0; i < arraysize ; i++)
  {
    converted = converted *10 + a[i] - '0';
  }

  printf("converted= %d", converted);   

  return 0;
}

Answer 2

Without using any existing libraries, you have to:

1. Convert character to numeric digit.
2. Combine digits to form a number.

Converting character to digit:

digit = character - '0';

Forming a number:

number = 0;
Loop:
number = number * 10 + digit;

Your function will have to check for '+' and '-' and other non-digits characters.

Answer 3

First of all this statement

 char a[5]='4534';

will not compile. I think you mean

 char a[5]="4534";
           ^^   ^^

To convert this string to a number is enough simple.

For example

int number = 0;

for ( const char *p = a; *p >= '0' && *p <= '9'; ++p )
{
    number = 10 * number + *p - '0';
}

Or you could skip leading white spaces.

For example

int number = 0;

const char *p = s;

while ( std::isspace( ( unsigned char )*p ) ) ++p;

for ( ; *p >= '0' && *p <= '9'; ++p )
{
    number = 10 * number + *p - '0';
}

If the string may contain sign '+' or '-' then you can check at first whether the first character is a sign.

Answer 4

I prefer to use the std::atoi() function to convert string into a number data type. In your example you have an non-zero terminated array "\0", so the function will not work with this code.

Consider to use zero terminated "strings", like:

char *a = "4534";

int mynumber = std::atoi( a );   // mynumber = 4534