how can I convert character to integer number - c++

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';
I want the number 4534 , How can I get it ?

Answer 1

Without using any existing libraries, you have to:
Convert character to numeric digit.
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 2

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 3

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 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