Code-Bin #1246

sdg
Posted by: asd | May 2, 2011 @ 5:33pm

C++ Code

// V �et�zci char num [20] je ulo�en v�raz sest�vaj�c� se ze dvou cel�ch ��sel odd�len�ch aritmetick�m oper�torem: +,-,* nebo /. Ze znak� p�ed oper�torem vytvo�te cel� ��slo A, ze znak� za oper�torem cel� ��slo B. V n�pov�d� jste zjistili �e k p�evodu m��ete vyu��t funkci int atoi( char *str) z knihovny stdlib.h. S takoto p�eveden�mi ��sly vykonejte p�edepsanou operaci a v�sledek vyp�ite do konzolov�ho okna. [5b]

// -------------------------------------

#include <string>
typedef enum e_oper{
	nic=0, plus, minus, krat, deleno
}xx;

//main
	char num[20]="3.141592*2.731";

char mul='*';
	char div='/';
	char sum='+';
	char sub='-';
	e_oper op=nic;

char pom[20];
	int x=0;
	while( x++<20 && op==nic) 
		if(num[x]==mul)		op=krat; 
		else if(num[x]==div) op=deleno; 
		else if(num[x]==sum) op=plus; 
		else if(num[x]==sub) op=minus; 
	
	for(int i=0; i<x; i++)
		pom[i] = num[i];
	pom[x] = 0;
	double a = atof(pom);
		
	for(int i=0; i<20; i++){
		pom[i] = num[x+i+1];
		if(num[i]==0) break;
	}
	double b = atof(pom);
	double c = 0;

switch (op){
		case plus:
			c=a+b;
			break;
		case minus:
			c=a-b;
			break;
		case krat:
			c=a*b;
			break;
		case deleno:
			c=a/b;
			break;
	}
	printf( "%f %c %f = %f\n", a, num[x], b, c );

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