9
In division 1 by 3, my program is printing the following:
number e' of 0.00
What is the error in the code?
#include <stdio.h>
int main(){
float numero;
numero = 1/3;
printf("o valor do numero e' de :%4.2f \n\n", numero );
return 0;
}
12
Because the code is dividing an integer by an integer.
You used a literal number which is an integer value. When you consider only integers, the division of 1 by 3 gives 0 the same. After the calculation results in zero, it is converted to float by the rule of casting automatic. But note that this casting only occurs with the result as a whole and not in each individual operand.
#include <stdio.h>
int main() {
printf("o valor do numero e': %4.2f", 1.0f/3.0f);
}
Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.
Using the numeric literal for the floating point type (1.0f for example), the division occurs in the correct way.
8
The problem is that the expression 1/3 is evaluated in the context of whole numbers. In an entire division, 1/3 == 0 (the expression is evaluated before being assigned to the variable numero).
If you use a value float in your division then you will have the value you expect, as in the example below:
#include <stdio.h>
int main(){
float numero;
numero = 1.0f/3;
printf("o valor do numero e':%4.2f \n\n", numero );
return 0;
}
2
#include <stdio.h>
int main(){
float numero,a,b;
a=1;
b=3;
numero = a/b;
printf("o valor do numero e':%.2f \n\n",numero);
return 0;
}
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– Maniero
Always prefer (*) to use
doubleinstead offloat. (*) except when the teacher keeps insisting after you explain to him the advantages ofdouble(is the only descantagem that does not make sense in "programetas").– pmg