CODEWITHSUNDEEP
c++
jam mal
jam mal

Reputation: 21

Converting int to float?

Whenever I try to compile I get

24 [Warning] converting to int from float

83 [Warning] converting to int from float 

int a , b = 8;
float c = 5.2;
float d = 8E3;
a = static_cast<float>(b) * c; // 24
cout << a << "\n";
cout << d << "\n";


int x, y, answer;
x = 7;
y = 9;
answer = 5;
answer *= (x + y);
cout << answer << "\n";


answer *= x + y;
cout << answer << "\n";


float m = 33.97;
answer += (x + y + m); // 83
cout << answer << "\n";

Any suggestions as to what I'm doing wrong?

Upvotes: 2

Views: 8011

Answers (3)

ypnos
ypnos

Reputation: 52317

Your question seems to be about conversion from float to int, not from int to float.

Basically you are doing nothing wrong. It is only a warning, as the value will be truncated (and maybe you don't expect that). To tell the compiler that you really want to get an int out of a float, you can make the cast explicit, like this:

a = static_cast<int>(static_cast<float>(b) * c);

Then it will not warn you anymore.

Upvotes: 2

erlc
erlc

Reputation: 680

Well you're just getting a warning since the compiler is changing a floating-point value to an integer, thus truncating the result.

int a;
float f = 3.2;
a = f; // a is 3, a trunctated 3.2

Upvotes: 1

kevintodisco
kevintodisco

Reputation: 5181

a = static_cast<float>(b) * c;

a is an int, and the right-hand side of the equation is the multiplication of two floats, which will result in an intermediate float value, which is then implicitly casted to an int, causing the warning you are seeing.

Also:

answer += (x + y + m);

answer is an int type, and so are x and y, but m is float, again causing the intermediate result of the right-hand side to be a float.

These conversions will cause truncation of the fractional values of the float results. You can get rid of the warnings by explicitly casting to an int:

a = static_cast<int>(static_cast<float>(b) * c);
answer += static_cast<int>(x + y + m);

Upvotes: 5

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