How do you convert char numbers to int numbers in C++?

Two ways you could do it:

1)

include <stdlib.h>

use the function strtol to convert strings to longs

long int strtol ( char * input, char* endOfNumber, int base)

Unfortunately this works on strings not chars so what you'd need to so is something like:

char myString[2];

long values[10];

myString[1] = 0;

for (int i = 0; i<10; i++) {

myString[0] = a[i];

values[i] = strtol(myString, NULL, 16);

}

The alternative method is to make your own function based on the ascii values of the char:

int hexCharToInt(char* input) {

if ((*input >=48) && (*input <= 57)) // 0 to 9

return (*input - 48);

if ((*input >= 65) && (*input <= 70)) // A to F

return (*input - 55);

if ((*input >= 97) && (*input <= 102)) // a to f

return (*input - 87);

return -1;

}

your code then becomes:

int values[10];

for (int i = 0; i<10; i++) {

values[i] = hexCharToInt(a[i]);

}

each and every char has a particular integer value assigned to it. examine the ASCII tables in case you pick for to correctly known. To get that integer value, all you do is solid the char to an int. To get it so the char A could be 0, only subtract int value for 'A' out of your char. So 'A' - 'A' = 0, 'B' - 'A' = a million, and so on. So working example: char c = 'B'; int i = (int)c - (int)'A'; printf("%i", i); //you need to get a million. easily, i found out you do no longer might desire to casting. The compiler does that for you.

int array[9];

for (int i=0;i<9;i++)

{

array[i]=a[i];

}

for (int i=0;i<9;i++)

{

cout<<hex<<array[i];

}