Expected } in C code - c

Question

So, keep having problems with this calculator I'm making (Still learning C), it's giving me an error at the very end of the code, 'Expected }'. But I have } at the end, Ive added another one to see if that would work, but then it says 'Expected while in do/while loop'.
#include <stdio.h>
#include <stdbool.h>
int main()
{
float number[100];
int operator = '0';
int doAgainAnswer = 0;
int doAgain;
float finished;
int error;
do{
if(number[1] == 0){
printf("Number 1: ");
scanf("%f", &number[1]);
} else if(number[1] != 0) {
printf("\n\nNumber 1: %.2f\n", number[1]);
printf("Number 2: ");
scanf("%f", &number[2]);
printf("\nOperator:\n1 Addition\n2 Subract\n3 Multiply\n4 Divide.\nChoice: ");
scanf("%d", &operator);
if(operator == 1){
finished = number[1] + number[2];
printf("\n\n%.2f \+ %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 2){
finished = number[1] - number[2];
printf("\n\n%.2f \- %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 3){
finished = number[1] * number[2];
printf("\n\n%.2f \* %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 4){
finished = number[1] / number[2];
printf("\n\n%.2f \/ %.2f is: %.2f", number[1], number[2], finished);
}
if(operator > 4){
printf("\n\nERROR: Invalid operation.\n\n ");
error = '1';
}
if(error != 1){
printf("\nContinue?\n1: Yes\n2: Yes and use answer as starting value.\n3: No\nAnswer: ");
} else if(error == 1){
printf("\nTry again? 1 Yes 3 No: ");
}
scanf("%d", &doAgainAnswer);
if(doAgainAnswer == 1) {
doAgain = 1;
} else
if(doAgainAnswer == 2){
doAgain = 1;
number[1] = finished;
} else
if(doAgainAnswer == 3){
doAgain = 0;
printf("Goodbye :(");
exit(0);
} else
if(doAgainAnswer > 4){
printf("\n\nERROR INVALID OPERATION.\n\n");
}
}while(doAgain == 1);
return 0;
}

Answer 1

It looks like this part of your code contains an else which is never closed, i.e.:
if(number[1] == 0)
{
printf("Number 1: ");
scanf("%f", &number[1]);
}
else if(number[1] != 0)
{
....
} <-- missing.
Whole code block, reformatted slightly differently:
#include <stdio.h>
#include <stdbool.h>
int main()
{
float number[100];
int operator = '0';
int doAgainAnswer = 0;
int doAgain;
float finished;
int error;
do
{
if(number[1] == 0)
{
printf("Number 1: ");
scanf("%f", &number[1]);
}
else if(number[1] != 0)
{
printf("\n\nNumber 1: %.2f\n", number[1]);
printf("Number 2: ");
scanf("%f", &number[2]);
printf("\nOperator:\n1 Addition\n2 Subract\n3 Multiply\n4 Divide.\nChoice: ");
scanf("%d", &operator);
if(operator == 1)
{
finished = number[1] + number[2];
printf("\n\n%.2f \+ %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 2)
{
finished = number[1] - number[2];
printf("\n\n%.2f \- %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 3)
{
finished = number[1] * number[2];
printf("\n\n%.2f \* %.2f is: %.2f", number[1], number[2], finished);
}
if(operator == 4)
{
finished = number[1] / number[2];
printf("\n\n%.2f \/ %.2f is: %.2f", number[1], number[2], finished);
}
if(operator > 4)
{
printf("\n\nERROR: Invalid operation.\n\n ");
error = '1';
}
if(error != 1)
{
printf("\nContinue?\n1: Yes\n2: Yes and use answer as starting value.\n3: No\nAnswer: ");
}
else if(error == 1)
{
printf("\nTry again? 1 Yes 3 No: ");
}
scanf("%d", &doAgainAnswer);
if(doAgainAnswer == 1)
{
doAgain = 1;
}
else if(doAgainAnswer == 2)
{
doAgain = 1;
number[1] = finished;
}
else if(doAgainAnswer == 3)
{
doAgain = 0;
printf("Goodbye :(");
exit(0);
}
else if(doAgainAnswer > 4)
{
printf("\n\nERROR INVALID OPERATION.\n\n");
}
}
}
while(doAgain == 1);
return 0;
}

Answer 2

else if(number[1] != 0) {
printf("\n\nNumber 1: %.2f\n", number[1]);
You never close this else if statement.

Answer 3

To prevent errors like this from happening, I suggest that you use a different kind of indentation.
If you write your code like this:
int main()
{
float number[100];
int operator = '0';
int doAgainAnswer = 0;
int doAgain;
float finished;
int error;
do
{
if(number[1] == 0)
{
printf("Number 1: ");
scanf("%f", &number[1]);
}
else if(number[1] != 0)
{
printf("\n\nNumber 1: %.2f\n", number[1]);
printf("Number 2: ");
scanf("%f", &number[2]);
printf("\nOperator:\n1 Addition\n2 Subract\n3 Multiply\n4 Divide.\nChoice: ");
scanf("%d", &operator);
You should notice instantly that there's a }missing after the else if statement.

Answer 4

Try having writing the {} in newlines and having them with no code in that line. it will help you get organized and find problems easier!

Answer 5

Proper indentation shows that you are missing a } to match your first else if {.
You really need to use a proper editor (one that at least does auto-indentation), use {} around your elses to avoid dangling them*, and use helper functions.
#include <stdio.h>
#include <stdbool.h>
int main()
{
...
do{
if(number[1] == 0){
printf("Number 1: ");
scanf("%f", &number[1]);
} else if(number[1] != 0) {
...
}while(doAgain == 1);
return 0;
}
*: seriously, always use braces unless it's short enough to go on a single line (else return, or similar). I found this in code at my last company:
if(DEBUG)
// print(foo);
if(something_important) {
...
}

Answer 6

Your code needs better indentation. With proper indentation it'd be easier to spot where the is (or is not) the missing bracket. You didn't close the outer if/else statement, right before the while condition:
#include <stdio.h>
#include <stdbool.h>
int main()
{
float number[100];
int operator = '0';
int doAgainAnswer = 0;
int doAgain;
float finished;
int error;
do {
if(number[1] == 0) {
printf("Number 1: ");
scanf("%f", &number[1]);
}
else if (number[1] != 0) {
printf("\n\nNumber 1: %.2f\n", number[1]);
printf("Number 2: ");
scanf("%f", &number[2]);
printf("\nOperator:\n1 Addition\n2 Subract\n3 Multiply\n4 Divide.\nChoice: ");
scanf("%d", &operator);
if (operator == 1) {
finished = number[1] + number[2];
printf("\n\n%.2f \+ %.2f is: %.2f", number[1], number[2], finished);
}
if (operator == 2) {
finished = number[1] - number[2];
printf("\n\n%.2f \- %.2f is: %.2f", number[1], number[2], finished);
}
if (operator == 3) {
finished = number[1] * number[2];
printf("\n\n%.2f \* %.2f is: %.2f", number[1], number[2], finished);
}
if (operator == 4) {
finished = number[1] / number[2];
printf("\n\n%.2f \/ %.2f is: %.2f", number[1], number[2], finished);
}
if (operator > 4) {
printf("\n\nERROR: Invalid operation.\n\n ");
error = '1';
}
if (error != 1) {
printf("\nContinue?\n1: Yes\n2: Yes and use answer as starting value.\n3: No\nAnswer: ");
}
else if (error == 1) {
printf("\nTry again? 1 Yes 3 No: ");
}
scanf("%d", &doAgainAnswer);
if (doAgainAnswer == 1) {
doAgain = 1;
}
else if(doAgainAnswer == 2) {
doAgain = 1;
number[1] = finished;
}
else if (doAgainAnswer == 3) {
doAgain = 0;
printf("Goodbye :(");
exit(0);
}
else if (doAgainAnswer > 4) {
printf("\n\nERROR INVALID OPERATION.\n\n");
}
/**
* Missing a bracket here, to close the if/else statement
*/
}
while (doAgain == 1);
return 0;
}
// Indentation should end at this level