编译时非法使用浮点错误

Question

I am getting an error as "illegal use of floating point" while running the program below.

How should I fix it.

#include<stdio.h>
#include<conio.h>
int main() {
   float x,op;
   printf("enter the value of x");
   scanf("%f",&x);
   op=(x^1/2+x^2/3+x^3/4)/(x^5/2+x^7/2);
   printf("Final Op is %f\n",op);
   return 0;
}

Answer 1

In C, ^ is used as the bitwise XOR operator. Each of the operands shall have integer type.

6.5.11 Bitwise exclusive OR operator:

2 Each of the operands shall have integer type.
4 The result of the ^ operator is the bitwise exclusive OR of the operands [...]

You need the standard library function powf :

float powf(float x, float y);  

Include math.h header in your program.

Also note that as pointed out by @Jens Gustedt 1/2 , 2/3 and 3/4 will all return 0 because they are integer division. You need to change them to 1.0/2 , 2.0/3 and 3.0/4 .

Answer 2

In C and related languages ^ is the bitwise XOR operator. For exponentiation you need pow() or powf() .

Change:

op=(x^1/2+x^2/3+x^3/4)/(x^5/2+x^7/2);

to:

op = (pow(x, 1./2) + pow(x, 2./3) + pow(x, 3./4)) / (pow(x, 5./2) + pow(x, 7./2));

Make sure you #include <math.h> .

Also you might want to get a good introductory book on C and read up on operators and math library functions.

#include <stdio.h>
#include <math.h>

int main()
{
    float x, op;
    printf("Enter the value of x: ");
    scanf("%f", &x);
    op = (pow(x, 1./2) + pow(x, 2./3) + pow(x, 3./4)) / (pow(x, 5./2) + pow(x, 7./2));
    printf("\nFinal op is %f\n", op);
    return 0;
}

LIVE DEMO

Answer 3

There are two problems with the op expression:

1. ^ is not the exponential operator in C, it is bitwise-XOR and required integer operands.
2. When both operands of / are integers, then an integer divide is performed, which truncates to the integer nearest zero. So for example 7/2 == 3 . You must make one or both operands floating point to perform floating point division such that 7.0/2.0 == 3.5

.

#include <math.h>

...
 op = ( pow( x, 1.0/2.0 ) +
        pow( x, 2.0/3.0 ) +
        pow( x, 3.0/4.0) ) / 
      ( pow( x, 5.0/2.0 ) + 
        pow( x, 7.0/2.0 ) ;

Note that the expression involves an implicit cast to float . In C89 the math functions are defined for double - so the cast cannot be avoided. C99 provides float variants; eg powf() , and C++ overloads them so that the type is determined by the operands; eg pow(7.0f/2.0f) .