pvjc26/du3/program.c

171 lines
5.3 KiB
C

#include <stdlib.h>
#include <stdbool.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <math.h>
#define SIZE 128
#define TEMPBUFFERSIZE 32
//kontrola, ze ci vycitane informacie su aritmeticky spravne
//VRACIA: int: -1 (CHYBA), 0 (False), 1 (True)
float IsArithCorrect(float num1, float num2, char op, float result)
{
float calcResult = 0;
if (op == '/' && num2 == 0) { return 0;}
switch (op)
{
case '-':
calcResult = num1 - num2;
break;
case '+':
calcResult = num1 + num2;
break;
case '*':
calcResult = num1 * num2;
break;
case '/':
calcResult = num1 / num2;
break;
}
calcResult = roundf(calcResult* 100.0f) / 100.0f;
result = roundf(result * 100.0f) / 100.0f;
if (calcResult == result)
{
return 1;
}
return 0;
}
int main()
{
//pomocne premeny
char line[SIZE], strBuffer[SIZE];
line[0] = '\0';
strBuffer[0] = '\0';
char* strtofEndPtr;
int LinesOnInputCount = 0;
while (fgets(line, SIZE, stdin) != NULL)
{
if (sscanf(line, "%127[^\n]", strBuffer) == 1)
{
bool isLineInvalid = false;
//docasne 'char' premeny
char num1Temp[TEMPBUFFERSIZE], num2Temp[TEMPBUFFERSIZE], resultTemp[TEMPBUFFERSIZE];
num1Temp[0] = '\0';
num2Temp[0] = '\0';
resultTemp[0] = '\0';
bool foundEqualSign = false;
char op1 = 'N'; //N = 'None/NIC'
//Ak je prazdny char na zaciatku, tak ocakavaj dalsi riadok
if (line[0] == ' ')
{
LinesOnInputCount += 1;
continue;
}
for (int i = 0; strBuffer[i] != '\0'; i++)
{
bool isNegativeSign = false;
if (isspace(strBuffer[i]) != 0)
{
continue;
}
if (strBuffer[i] == '-')
{
//kontrola - zistenie, ze ci znak '-' symbolizuje zapornu hodnotu (-456) alebo arit. operaciu minus
//zistujeme to na zaklade toho, kde sme v riadku (ak sme na zaciatku num1, num2 alebo vyslednehoCisla, tak sa jedna o minusove znamienko)
if (op1 == 'N' && strlen(num1Temp) == 0)
{
isNegativeSign = true;
}
else if (op1 != 'N' && !foundEqualSign && strlen(num2Temp) == 0)
{
isNegativeSign = true;
}
else if (foundEqualSign && strlen(resultTemp) == 0)
{
isNegativeSign = true;
}
}
if (isdigit(strBuffer[i]) != 0 || strBuffer[i] == '.' || isNegativeSign)
{
//cislo sa bude zapisovat podla toho, kde sme pri citani riadku...
//ak nenasiel ani arit. znamienko = cislo patri ku prvemu operandu (num1)
//ak nasiel arit. znamienko ale NIE este symbol '=', tak cislo patri ku druhemu operandu (num2)
//inak ak ma vsetko, tak zvysne nacitane cisla z riadku mozu byt vyluc. metodou len sucastou vysledku
if (op1 == 'N')
{
strncat(num1Temp, &strBuffer[i], 1);
}
else if (!foundEqualSign)
{
strncat(num2Temp, &strBuffer[i], 1);
}
else
{
strncat(resultTemp, &strBuffer[i], 1);
}
}
else if (strBuffer[i] == '=')
{
foundEqualSign = true;
continue;
}
else if (strBuffer[i] == '+' || strBuffer[i] == '-' || strBuffer[i] == '/' || strBuffer[i] == '*')
{
//predbezna kontrola, ak uz operator mame tak vyhod chybu pri druhom pokuse o nacitanie
//taktiez, ak sa nasiel minusovy znak tu tak to moze byt len arit. znamienko
if (op1 != 'N')
{
isLineInvalid = true;
}
op1 = strBuffer[i];
}
else
{
isLineInvalid = true;
break;
}
}
float num1 = strtof(num1Temp, &strtofEndPtr);
float num2 = strtof(num2Temp, &strtofEndPtr);
float resultNum = strtof(resultTemp, &strtofEndPtr);
//** konecna faza = vypisovanie vysledkov
//ak systematicka chyba
if (isLineInvalid || op1 == 'N' || IsArithCorrect(num1, num2, op1, resultNum) == -1)
{
printf("CHYBA\n");
}
else if (IsArithCorrect(num1, num2, op1, resultNum) == 0)
{
printf("ZLE\n");
}
else
{
for (int i = 0; i <= LinesOnInputCount; i++)
{
printf("OK\n");
}
}
}
}
return 0;
}