cout to printf problems

i m trying to port something of cout to printf this is the cout:

Quote:

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int y = 1;
int x =1;
cout<<"schrijf 2 getallen voor te vermenigvuldigen: \n";
cout<<"het produkt van de 2 getallen is :";
cout<<x*y;

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and this is the not working printf version :

Quote:

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int y = 1;
int x =1;
double a = (x*y);


printf("schrijf 2 getallen voor te vermenigvuldigen: \n");

printf("het produkt van de 2 getallen is : %a ");

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i have got some reasons to make it printf so how do i have to do that?

Code:

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int y = 1;
int x =1;
printf("schrijf 2 getallen voor te vermenigvuldigen: \n");
printf("het produkt van de 2 getallen is :%d", x*y);

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Code:

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printf("The product of the two integers is: %f ", x*y);

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But the product doesn't have to be a double, int*int=int.

(thanks RichieT)

the compille had a warning that's says that it better will be a double so.....;)

I've seen this problem many times.
Somehow, people think that printf acts as it does in PHP.

I made many such mistakes when I was learning C++...

Quote:

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the compille had a warning that's says that it better will be a double so.....

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You compiled my version, or twomers' version? Incidentally, you might want to read about printf.

Quote:

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Originally Posted by twomers

Code:

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printf("The product of the two integers is: %f ", x*y);

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But the product doesn't have to be a double, int*int=int.

(thanks RichieT)

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That won't work right. Printf will expect the second peramitter to be a double, but you're sending it an int. The compiler won't complain because it's a veradic function. Yet when you try to print it out it will try to read the int as if it were a double. It might crash, look beyond the input value in memmory, because ints are generally smaller than doubles (not sure about this). If it doesn't, it will print out the bitwize value as if it was a double. And since 1 is not encoded in an int the same whay as it is in a double, the value will be different.

You would need to do this:

Code:

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printf("The product of the two integers is: %f ", (double)(x*y));

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If x or y are large enough, then the result would overflow an integer. Maybe that is why the double is being used?

Try setting x and y to INT_MAX and you'll see what I mean. (double)(x * y) won't work in that scenario.

Code:

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printf("het produkt van de 2 getallen is : %.f\n", (double) x * (double) y);

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Note the use of "%.f" so the output "appears" to be an integer.

Een enkel gedachte!

i used the version of laserlight and it works

thnx to all;)

You're right. I got kinda confused with the whole -

Quote:

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double a = (x*y);

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thing, and the printf, so instead of printf'in "%f", a, I did "%f", x*y, if ya get me. I should have had a there instead of x*y. My bad.