How can I take several figures as once from my terminal - c

Question

I'm trying to take several figures from my terminal to make my calculator work correctly. I would have to take every figures before a space : eg. 160 + 23 I have to get 160 for my first variable, "+" for my second one and 23 for my last one.
Here is the code :
`
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUFFER 25
void calculatrice(double,char,double);
int main(){
char *str;
printf("************************************************\n");
printf("ecrire sous la forme x + y avec les espaces pour avoir un resultat\n");
printf("quit pour quitter\n");
printf("************************************************\n");
char *text = calloc(1,1), buffer[BUFFER];
while( fgets(buffer, BUFFER , stdin)){
if (!strncmp(buffer, "quit",strlen("quit"))){
break;
}
if (!strncmp(buffer, "flush",strlen("flush"))){
system("clear");
}
text = realloc( text, strlen(text)+1+strlen(buffer) );
if( !text ){
strcat( text, buffer);
printf("%s\n", buffer);
}
double nb1 = strtol(&buffer[0], &str, 10);
double nb2 = strtol(&buffer[3], &str, 10);
calculatrice(nb1, buffer[2], nb2);
}
return 0;
}
void calculatrice(double value_1, char operator, double value_2){
double result = 0;
switch (operator)
{
case '+':
result = value_1 + value_2;
printf("resultat : %f\n",result);
break;
case '-':
result = value_1 - value_2;
printf("resultat : %f\n",result);
break;
case '*':
result = value_1 * value_2;
printf("resultat : %f\n",result);
break;
case '/':
result = value_1 / value_2;
printf("resultat : %f\n",result);
break;
case '%':
result = (int)value_1 % (int)value_2;
printf("resultat : %f\n",result);
break;
default:
break;
}
}
`
I tried to make loops and conditions which triggers when there is a space to get each part necessary for my calculator, but the result was completely wrong.

Answer 1

Given the user input of
160 + 23
then
&buffer[0] will be the string "160 + 23\n",
&buffer[3] will be the string " + 23\n", and
buffer[2] will be the character '0'.
strtol converts the initial part of a string to a long. With base-10, the string may begin with an arbitrary amount of whitespace, and then an optional sign character of '+' or '-'.
For "160 + 23\n", the converted value will be 160 as it stops on the first non-digit character encountered (a ' ' after '0').
For " + 23\n", the converted value will be 0, after skipping the initial whitespace, reading the sign character '+', and stopping on the first non-digit character encountered (a ' ' after '+').
These resulting values are then implicitly converted to double during the assignment.
Note, to convert a string directly to a double, use strtod.
The effect is that, with the given user input, calculatrice is called as:
calculatrice(160.0, '0', 0.0);
There is no match for '0' in the switch, and thus the function quietly returns.
You must find the correct string tokens more carefully, but how this is done depends greatly on the requirements of your program. Two interesting functions in the standard library are:
strtok can be used to break a string into tokens, based on a set of delimiting characters.
sscanf can be use to convert well formatted strings into data.
Here is a cursory example of using sscanf.
#include <stdio.h>
#include <stdlib.h>
double calculatrice(double, char, double);
int main(void)
{
puts("************************************************");
puts("ecrire sous la forme x + y avec les espaces pour avoir un resultat");
puts("quit pour quitter");
puts("************************************************");
char buffer[128];
while (fgets(buffer, sizeof buffer, stdin)) {
if ('q' == *buffer)
break;
if ('f' == *buffer) {
system("clear");
continue;
}
char op;
double lhs, rhs;
if (3 == sscanf(buffer, "%lf %c%lf", &lhs, &op, &rhs))
printf("Result: %f\n", calculatrice(lhs, op, rhs));
}
}
double calculatrice(double a, char operator, double b)
{
switch (operator) {
case '+':
return a + b;
case '-':
return a - b;
case '*':
return a * b;
case '/':
if (0 == b) {
fprintf(stderr, "Division by zero detected.\n");
break;
}
return a / b;
case '%':
if (0 == (int) b) {
fprintf(stderr, "Division by zero detected.\n");
break;
}
return (int) a % (int) b;
default:
fprintf(stderr, "Invalid operator [%c].\n", operator);
break;
}
return 0.0;
}
************************************************
ecrire sous la forme x + y avec les espaces pour avoir un resultat
quit pour quitter
************************************************
160 + 23
Result: 183.000000
The purpose of text here is unclear, but the use of
if (!text) {
strcat(text, buffer);
/* ... */
is surely wrong, as it guarantees Undefined Behaviour by passing the NULL pointer value to strcat. Otherwise, text is unused until this failure state.