Why am I getting 0.00 when I enter alphabetical character? - c

Question

The following is the code that I wrote. I feel like I have missed something with the pointer or might have made mistake in the calculation. But as for as I see, everything seems good for me but when I enter, for example: hello or anything else, it gives me 0.00 as output.
The below is correct
Enter a number:
6
Your number is: 6.00
But this why?
Enter a number:
h
Your number is: 0.00
Following is the complete code:
#include <stdio.h>
#include <stdlib.h>
#define SIZE 250
float myAtof(char *string, char *error);
int main() {
char string[SIZE]; // Array declaration.
float fnum1;
char errorState=0;
printf("Enter a number:\n");
gets(string);
fnum1=myAtof(string,&errorState);
if (errorState == 0) {
printf("Your number is: %.2f \n", fnum1);
} else if (errorState == 1) {
printf("Error has been occurred due to inappropriate input!\n");
}
return 0;
}
float myAtof(char* string, char* error) {
*error != '1';
float num= 0;
float decimalFlag = 1;
int decimalNumberFound = 0;
for (int i = 0; string[i]!= '\0'; i++) {
char ch = string[i];
if (ch != '.' && (ch < '0' || ch >'9')) {
error ='1';
return 0;
}
if (decimalNumberFound == 1 && ch == '.') {
error = '1';
return 0;
} else {
num = num* 10 + (ch - '0');
if (decimalNumberFound == 1)
decimalFlag *= 10;
}
}
num = num / decimalFlag;
return num;
}

Answer 1

Since your myAtof routine has a parameter for reporting an error and your main routine tests the error-reporting variable, errorState, presumably you expect the program to print an error message when you enter “h” or other non-numerals, and your question is why it prints “0.00” rather than the error message.
In the function declared with float myAtof(char* string, char* error), error is a pointer, and *error is the thing it points to. So, when the code executes error = '1';, that attempts to set the pointer to '1'. It does not change the thing it points to.
When you compiled this, the compiler issued a warning message, something like “warning: incompatible integer to pointer conversion assigning to 'char *' from 'int'”. You should not have ignored that message.
The function also contains the statement *error != '1';. That statement does nothing. It says to compare the value of *error to '1' to see if they are unequal. The result of the != operator is 1 or 0 indicating whether the values are unequal or not. But the statement does nothing with that result. It is just discarded, so the statement has no effect.
Nothing in your function changes the value of *error, so the thing it points to, errorState in main, is never changed. So main does not see that an error has occurred. It evaluates errorState==0 as true and executes printf("Your number is: %.2f \n", fnum1);.
To fix this, delete the statement *error != '1'; and change the two error = '1'; statements to *error = '1';.
Additionally, change '1' to 1. Setting *error to '1' sets it to the code for the character “1”. It does not set it to the value 1, which is what errorState == 1 tests for. Typically, an error status would be either 0 or 1, not 0 or '1'.
Also change else if (errorState == 1) to else. When you have an if and else that are intended to completely select between some condition X and its alternative, you do not need a second test. With your code if (X) { … } else if (Y) { … }, the first { … } is executed if X is true, the second { … } is executed if X is false and Y is true, and neither is executed if X is false and Y is false. But it should never be the case that X is false and Y is false—you want the program always to either have a number or have an error status. There should be only two possibilities, not three. So use just else, not else if.
(Conceptually, it is a bug for the program to have errorState be something other than 0 or 1, so the program would never have both X false and Y false if it were working, and the else if would be okay. But you did have a bug, and that would cause neither { … } to be executed. Designing the code to use else instead of an unnecessary else if makes it a little more resistant to misbehaving when there is a bug. Designing programs to be robust in this way is useful.)
Also pay attention to warning messages from your compiler. Preferably, tell the compiler to treat all warnings as errors, so that your program will not compile while warning messages remain. If you are using GCC or Clang, use the -Werror switch. If you are using Microsoft Visual C++, use /WX.