Why does the number get ignored? - c

Question

I want to write a very simple calculator in C.
At the start, the variable for the output is 0 and every calculation adjusts the value.
For example if I type -a 5 at the program start the output is 5, if I write -a 5 -s 5 the output is 0.
If I don't choose a or s it will just add all values to the output.
And if I type something like -a 10 -s 5 10 25, the 10 and 25 also should be add to the output.
This is my code so far:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
int result = 0;
for (int i = 1; i < argc - 1; i++) {
for (int j = i + 1; j < i + 2; j++) {
int value = atoi(argv[j]);
if (strcmp(argv[i], "-a") == 0) {
result += value;
} else
if (strcmp(argv[i], "-s") == 0) {
result -= value;
} else {
result += value;
}
}
}
printf("%d\n", result);
return 0;
}
All works fine, but not when I just type in some numbers without -s or -a.
When I start the program for example with 5 10 25, it will ignore the first number and the output will be 35. I don't really know, how to fix this.

Answer 1

Problems:
The first argument will never be treated as value inside of your program because j in int value = atoi(argv[j]); will ever start with the value of 2, not 1. j is declared and initialized with int j = i + 1; (j gets initialized with the value of i plus one) and since i starts with the value of 1 (int i = 1), j will start with a value of 2.
The inner loop isn't needed at all and literally only mess things up as it is even the source of your main issue already and makes your code harder to read. I highly recommend you to omit it.
For value: Declaring a variable inside a loop isn't a good practice BTW because the variable is declared new at each iteration. A compiler can optimize this but just avoid it.
Also a problem is that you convert the string arguments of -a and -s into an int value with this because you use the conversion before checking the values of the arguments. Note that used in the right way (when the conversion is only done based on a value argument) we don't need the variable value at all.
You can simplify the code like that:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main (int argc, char *argv[]) {
int result = 0;
for (int i = 1; i < argc; i++) {
if ( strcmp(argv[i], "-a") == 0 && i + 1 < argc && strcmp("-s", argv[i+1]) && strcmp("-a", argv[i+1]) ) {
result += atoi(argv[i+1]);
i++; // We use the next argument already here, so go forward.
}
else if ( strcmp(argv[i], "-s") == 0 && i + 1 < argc && strcmp("-s", argv[i+1]) && strcmp("-a", argv[i+1]) ) {
result -= atoi(argv[i+1]);
i++; // We use the next argument already here, so go forward.
}
else {
result += atoi(argv[i]);
}
}
printf("%d\n", result);
return 0;
}
Execution
Your example:
./calculator -a 10 -s 5 10 25
40
My example:
./calculator 10 34 -a 6 -s 4 25 -a 19 5 -s 24
71
Have fun at proof if the calculation is correct. ;-)
Or just try it online.

Answer 2

The problem with your code is that argv[1] is only tested to be either -a or -s but there is no code that converts argv[1] to a number and adds it to the result.
You need to handle all 3 cases in the loop, i.e. case 1 "-a", case 2 "-s" and case 3 "a number".
This could be like:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
int result = 0;
int i = 1;
while (i < argc) {
if (strcmp(argv[i], "-a") == 0) {
// Case 1
++i; // Increment i to move to next argument
if (i == argc) break; // Check that there is a valid argument
int value = atoi(argv[i]);
result += value;
} else
if (strcmp(argv[i], "-s") == 0) {
// Case 2
++i;
if (i == argc) break;
int value = atoi(argv[i]);
result -= value;
} else {
// Case 3 Current argument is (expected to be) a number
int value = atoi(argv[i]);
result += value;
}
++i; // Move to next argument
}
printf("%d\n", result);
return 0;
}
The above code uses atoi like the code in the question. A down-side of atoi is that there is no input validation, i.e. it's not validated that the string is actually a number.
For better input validation consider using strtol instead of atoi.

Answer 3

Your program does not seem to implement a solution to the problem:
You should keep track of the current mode ie: adding or subtracting and handle each argument accordingly (resetting the mode after each number handled).
Your nested loop ignores the first of 2 arguments. It also ignores the argument if only one is provided.
Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
int result = 0;
int mode = 1;
for (int i = 1; i < argc; i++) {
if (!strcmp(argv[i], "-a")) {
mode = 1;
} else
if (!strcmp(argv[i], "-s")) {
mode = -1;
} else {
char *p;
int value = strtol(argv[i], &p, 0);
if (p == argv[i] || *p != '\0') {
printf("invalid argument: %s\n", argv[i]);
} else {
result += mode * value;
}
mode = 1; // only subtract a single argument
}
}
printf("%d\n", result);
return 0;
}
Note that the above program detects invalid input but does not detect nor handle arithmetic overflow.