C programming : String calculation Using ASCII code [closed] - c

Question

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 6 years ago.
Improve this question
If I get the numbers in character string like this and store them in array x[100] and y[100],
x[100] '1' '0' '0' '0' '\0'
y[100] '9' '9' '\0'
The added result should be stored as string, too.
result[100] '1' '0' '9' '9' '\0'
I have tried to use ASCII code to solve this problem, for example:
char number[2] = {'1','2'};
char result;
result = (number[0] - '0') + (number[1] - '0') + '0';
But I am having trouble adjusting digits of x[100] and y[100].
I really need your help :(

Answer 1

You may insist on avoiding the library functions to convert strings to numbers and vice versa, as mentioned by #sjsam.
If you think about it, numbers really start at the LSD (least significant digit, not acid), or on the right side. So start your loop at the last character of each array before the closing \0 and iterate backwards. It won't be trivial, but that's how atoi() works too.
It's rather unfortunate that the Romans started writing from left to right, but that's the root cause of this problem. Otherwise integration of right-to-left Arabic numbers would have been much easier.
Discussing the idiocy of Roman numerals is out of the scope of this answer, let's suffice it to say that their lack of zero prevented any of their programs to finish with a successful exit status, which in turn led to the collapse of the Roman Empire.

Answer 2

Instead of adding byte by byte you may use the functions atoi to convert a string to integer and once you calculate the sum, you may use itoa/snprintf to convert the sum(integer) to its string representation.
See an example below :
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
char buffer[100];
int no1,no2,no3;
char number1[] = {'1','0','0','0','\0'};
char number2[] = {'9','0','0','\0'};
/* Converting strings to respective integers */
no1=atoi(number1);
no2=atoi(number2);
no3=no1+no2;
/* Convert integer to a null terminated string
* You could also use itoa(no3,buffer,10);
*/
snprintf(buffer,99,"%d",no3);
printf("no1 : %d\n",no1);
printf("no2 : %d\n",no2);
printf("no3 : %s\n",buffer); // printing the string
return 0;
}
itoa is not supported by some compilers, so the best bet is to use snprintf.
Output
Here buffer is a null terminated array
{'1','9','0','0','\0'}
which we have printed using the %s format specifier.
References
itoa manaul.
snprintf manual.
Note
I have set the buffer size to 100 for this example. However, the maximum value of number of bytes that buffer can hold is the depends upon the maximum value an integer(signed in your case) can contain in your system. (Thanks #BLUEPIXY for reminding this stuff. :) )

Answer 3

Sum ASCII numbers while avoiding converting to binary/decimal:
#include <stdio.h>
#include <string.h>
#include <libc.h>
char *sum(char *x, char *y) {
size_t x_idx = strlen(x);
size_t y_idx = strlen(y);
size_t z_idx = MAX(x_idx, y_idx) + 1;
char *z = malloc(z_idx + 1);
char carry = '0';
while (x_idx > 0 || y_idx > 0 || carry == '1') {
char digit = carry;
if (x_idx > 0 && y_idx > 0) {
digit = (x[--x_idx] + y[--y_idx] + carry) % '0' + '0';
} else if (x_idx > 0) {
digit = (x[--x_idx] + carry) % '0' + '0';
} else if (y_idx > 0) {
digit = (y[--y_idx] + carry) % '0' + '0';
}
carry = (digit > '9') ? '1' : '0';
if (carry == '1') {
digit -= 10;
}
z[--z_idx] = digit;
}
while (z_idx > 0) {
z[--z_idx] = ' '; // pad for now; for production, shift array
}
return z;
}
int main(int argc, char* argv[]) {
char *x = argv[1];
char *y = argv[2];
char *z = sum(x, y);
printf("%s + %s = %s\n", x, y, z);
free(z);
}
USAGE
> ./a.out 1000 99
1000 + 99 = 1099
>
> ./a.out 999 999
999 + 999 = 1998
>

Answer 4

#include <stdio.h>
#include <string.h>
int main(void) {
char x[100] = "1000";
char y[100] = "99";
char result[100+1] = " ";
int x_i = strlen(x)-1;
int y_i = strlen(y)-1;
int r_i = 1 + ((x_i > y_i) ? x_i + 1 : y_i + 1);
int carray = 0, sum;
result[r_i] = 0;
while(--r_i>0){
if(x_i >= 0 && y_i >= 0)
sum = x[x_i--] - '0' + y[y_i--] - '0' + carray;
//else if(x_i < 0 && y_i < 0)
// sum = carray;
else if(y_i < 0)// && x_i >= 0){
sum = x[x_i--] - '0' + carray;
else// if (x_i < 0 && y_i >= 0){
sum = y[y_i--] - '0' + carray;
carray = sum > 9;
result[r_i] = sum % 10 + '0';
}
if(carray)
result[0] = '1';
printf("%s\n", result);
return 0;
}

Answer 5

Here's some code that should give you a good idea.
This begins to handle carrying, but doesn't cover all cases.
It should be a good start.
#include <stdio.h>
#include <string.h>
int main(void) {
char y[100] = "1032";
char x[100] = "2399";
int carry = 0;
char* b = (strlen(x) > strlen(y))? x : y;
char* s = (strlen(x) <= strlen(y))? x : y;
for(int i=strlen(s)-1, j=strlen(b)-1; i>=0; --i,--j)
{
b[j] = (b[j]+s[i]+carry-'0');
carry = 0;
if (b[j] > '9')
{
b[j] = (b[j]-'0')%10+'0';
carry = 1;
}
}
puts(b);
return 0;
}