Convert String to Integer without library function in C [closed] - c

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I have to write a program that converts an user input (which is a string) to an Integer. In the same time it should check, if the user input really is a number.
And also everything in one method.
and NO LIBRARY FUNCTIONS allowed.
I can't figure any idea how to do it. All I got for the beginning is just this pathetic structure
#include <stdio.h>
void main()
{
char input[100];
int i;
int sum = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%c, &input");
for (i = 0; i < 100; i++)
{
}
}
I appreciate any help, thanks

Answer 1

The conversion is the easy part...
But if you must not use library functions,
there is only one way to take a string, and that is argv;
there is only one way to give an integer, and that is the exit code of the program.
So, without much ado:
int main( int argc, char * argv[] )
{
int rc = 0;
if ( argc == 2 ) // one, and only one parameter given
{
unsigned i = 0;
// C guarantees that '0'-'9' have consecutive values
while ( argv[1][i] >= '0' && argv[1][i] <= '9' )
{
rc *= 10;
rc += argv[1][i] - '0';
++i;
}
}
return rc;
}
I did not implement checking for '+' or '-', and did not come up with a way to signal "input is not a number". I also just stop parsing at the first non-digit. All this could probably be improved upon, but this should give you an idea of how to work around the "no library functions" restriction.
(Since this sounds like a homework, you should have to write some code of your own. I already gave you three big spoons of helping regarding argv, the '0'-'9', and the conversion itself.)
Call as:
<program name> <value>
(E.g. ./myprogram 28)
Check return code with (for Linux shell):
echo $?
On Windows it's something about echo %ERRORLEVEL% or somesuch... perhaps a helpful Windows user will drop a comment about this.
Source for the "'0'-'9' consecutive" claim: ISO/IEC 9899:1999 5.2.1 Character sets, paragraph 3:
In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.
I'm sure this is preserved in C11, but I only have the older C99 paper available.

Answer 2

Take hightes digit and add it to number, multiply the number by 10 and add the next digit. And so on:
#include <stdio.h> // scanf, printf
void main()
{
char input[100];
printf("Type a String which will be converted to an Integer: ");
scanf("%s", input);
int number = 0;
int neg = input[0] == '-';
int i = neg ? 1 : 0;
while ( input[i] >= '0' && input[i] <= '9' )
{
number *= 10; // multiply number by 10
number += input[i] - '0'; // convet ASCII '0'..'9' to digit 0..9 and add it to number
i ++; // step one digit forward
}
if ( neg )
number *= -1;
printf( "string %s -> number %d", input, number );
}
input[i] - '0' works, because ASCII characters '0'..'9' have ascending ASCII codes from 48 to 57.

Answer 3

So basically you want to know how something like the standard library atoi works. In order to do this, you need to consider how strings represent numbers.
Basically, a string (that represents a number) is a list o digits from 0 to 9. The string abcd (where a, b, c, d are placeholders for any digit) represents the number a*10 ^ 3 + b*10^2 + c * 10 + d (considering base 10 here, similar for other bases). So basically you need to decompose the string as shown above and perform the required arhitmetic operations:
// s - the string to convert
int result = 0;
for (int index = 0; index < strlen(s); index++) {
result = result * 10 + s[index] - '0';
}
The operation s[index] - '0' converts the character that represent a digit to its value.

Answer 4

// the function returns true for success , and false for failure
// the result is stored in result parameter
// nb: overflow not handled
int charToInt(char *buff,int *result){
*result=0;
char c;
while(c=*buff++){
if((c < '0') || (c >'9')) // accept only digits;
return 0;
*result *= 10;
*result += c-'0';
}
return 1;
}

Answer 5

Lot of things which are missed. Firstly taking a string in is done by scanf("%s",input); By the way in which you are receiving it, it only stores a character, secondly run the loop till the length of the string recieved. Check the below code.
#include <stdio.h>
#include<string.h>
void main()
{
char input[100];
int i;
int sum = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%s", input);
for (i = 0; i < strlen(input); i++)
{
if(input[i]>=48 && input[i]<=57)
{
//do something, it is a digit
printf("%d",input[i]-48);
//48 is ascii value of 0
}
}

Answer 6

Try it:
#include <stdio.h>
void main()
{
char input[100];
int i,j;
int val = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%s",input);
for(j=0; input[j] != '\0'; j++); // find string size
for (i = 0; i < j; i++)
{
val = val * 10 + input[i] - 48;
}
}

Answer 7

If you want your code to be portable to systems that don't use ASCII, you'll have to loop over your char array and compare each individual character in the source against each possible number character, like so:
int digit;
switch(arr[i]) {
case '0':
digit=0; break;
case '1':
digit=1; break;
// etc
default:
// error handling
}
Then, add the digit to your result variable (after multiplying it by 10).
If you can assume ASCII, you can replace the whole switch statement by this:
if(isdigit(arr[i])) {
digit=arr[i] - '0';
} else {
// error handling
}
This works because in the ASCII table, all digits are found in a single range, in ascending order. By subtracting the ordinal value of the zero character, you get the value of that digit. By adding the isdigit() macro, you additionally ensure that only digit characters are converted in this manner.