Thread: Promoting char to int

Hey,

I'm reading in a bunch of numbers into a char array, and then I want to perform mathematical operations on these numbers.
I've read the FAQ > Explanations of... > Definition of EOF and how to use it effectively where they talk about the promotion, but I'm still not sure if I can do what I want to do. Or maybe there's another way.
say I have and input like so:
126

I read it in one character at a time. so a string would contain:
1 2 6

Now I want to assign the value 126 to an int. In the code count represents the number of characters read in. I can't read in the characters differently because the file contains letters and numbers and the reading function reads one char at a time.

Code:

int ConvertNum(FILE *inp, char *string, int count)
{
	int i /*position*/, j =0 /*power*/, num =0, temp;

	for (i=(count-1); i >=0; i--)
	{
		temp = string[i];
		num = num + temp * (10^j);
		j++;

		printf("temp = %d, string[i] = %c  \n", temp, string[i]);
	}

	return num;
}

thanks.

The atoi() function turns a string made of number characters [0-9] into an integer. That what you want?

Originally Posted by jim mcnamara

The atoi() function

That's exactly what I want.
Thanks,

A.

Just my 2 cents. Not really sure if this is what u r looking for.

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
	char arr[] = {'1','2','3'};
	
	char num[10];

	int i,number;
	
	strncpy(num,arr,3);
		
	number=atoi(num);
	printf("Number: %d\n",number);

	
	return 0;
}

$ ./test1.exe
Number: 123

Code:

	num = num + temp * (10^j);

^ is the bitwise XOR operator. You are looking for the pow function.

Thanks,

Got it working,

A.