COUT Convert to PRINTF

[Shodow]

can u help me convert the cout to printf..i dont have a clue how..
tnx

#include<iostream.h>
#include<string.h>
#include<conio.h>
int findSimilar(char[],char[]);
void checkValue(int,int);
int main(){
	char boy[50];
	char girl[50];
	int similarInBoy=0;
	int similarInGirl=0;
	int total;
	clrscr();
	cout<<"Enter Boy's Name:  ";
	cin.getline(boy,50);

	cout<<"Enter Girl's Name: ";
	cin.getline(girl,50);

	similarInBoy=findSimilar(boy,girl);
	similarInGirl=findSimilar(girl,boy);
	total=similarInGirl+similarInBoy;

	cout<<"Boy: "<<similarInBoy<<",";
	checkValue(similarInBoy,1);
	cout<<"\n";

	cout<<"Girl: "<<similarInGirl<<",";
	checkValue(similarInGirl,2);
	cout<<"\n";

	cout<<"Total: "<<total<<",";
	checkValue(total,3);
	cout<<"\n";
	getch();
	return 0;
}

void checkValue(int similar,int gender){

	if(gender==1){
		if(similar>6)
			similar%=6;
		switch(similar){
		case 1:
			cout<<"Friend";
			break;
		case 2:
			cout<<"Love";
			break;
		case 3:
			cout<<"Affair";
			break;
		case 4:
			cout<<"Married";
			break;
		case 5:
			cout<<"Enemy";
			break;
		case 6:
			cout<<"Sweetheart";
			break;

		}
	}
	else if(gender==2){
		if(similar>6)
			similar%=6;
		switch(similar){
		case 1:
			cout<<"Friend";
			break;
		case 2:
			cout<<"Love";
			break;
		case 3:
			cout<<"Affair";
			break;
		case 4:
			cout<<"Married";
			break;
		case 5:
			cout<<"Enemy";
			break;
		case 6:
			cout<<"Sweetheart";
			break;
		}
	}
	else{
		if(similar>12)
			similar%=12;
		switch(similar){
			case 1:
			cout<<"Friends";
			break;
		case 2:
			cout<<"Lovers";
			break;
		case 3:
			cout<<"Affair";
			break;
		case 4:
			cout<<"Marriage";
			break;
		case 5:
			cout<<"Enemies";
			break;
		case 6:
			cout<<"Sweethears";
			break;
		}
	}
}


int findSimilar(char firstName[],char secondName[]){
	int len;
	int len2;
	int similar=0;
	char ch1,ch2;
	char* first;
	char *second;

	first=strupr(firstName);
	second=strupr(secondName);
	len=strlen(firstName);
	len2=strlen(secondName);

	for(int ctr=0;ctr<len;ctr++){
		ch1=first[ctr];
		for(int ctr2=0;ctr2<len2;ctr2++){
			ch2=second[ctr2];
			if((ch1!=' ')&&(ch1==ch2)){
				similar++;
				break;
			}
		}
	}
	return similar;
	}

[WaltP]

Look up the format of printf(). Use that info to do your conversion.

[Shodow]

my bad help me convert cout to printf and cin to scanf.. help me pls asap

[diwakar wagle]

my bad help me convert cout to printf and cin to scanf.. help me pls asap

here you go:

int printf(const char *format, ...)

But you have to include header file "stdio.h"("cstdio.h" if you're compiling as c++) in your code to use printf or scanf.
simple example of printf:

printf("Hello, world!\n")

Here's more about printf and scanf .

Use fgets instead , since scanf is not considered safe .

[Narue]

Use fgets instead , since scanf is not considered safe.

scanf() is perfectly safe when used correctly. The problem is things like %s without a field width where the programmer has no clue that it's a buffer overflow waiting to happen.

If scanf() is so unsafe, then why is the recommended alternative a combination of fgets() and sscanf()? The only major difference between scanf() and sscanf() is the source of characters. Further, if you read those links, you'll find that the "problems" with scanf() have to do with variations of the following:Not understanding how scanf() manages whitespace
Not understanding how the %s specifier works
While I agree that an intuitive interface and semantics are both important, ignorance of scanf() isn't the fault of scanf(). Were I in a perverse mood, I'd argue that fgets() also has problems because beginners are often confused about how newlines are managed.

[Shodow]

can u help me with scanf? cuz its must be printf and scanf..pls

how to convert this in scanf?

cin.getline(boy,50);

[Narue]

can u help me with scanf? cuz its must be printf and scanf..pls

how to convert this in scanf?

cin.getline(boy,50);

A C programmer would prefer fgets() for string input, but since this isn't an immediately obvious question about scanf(), I'll answer it. The closest equivalent to such a getline() call with scanf() is:

scanf("%49[^\n]", boy);