c - expected expression before 'long' -

i'm new programming, , i'm using book me write code solve ln(1+x) after user inputs "x". syntax way book has in example keep getting error: expected expression before 'long' on line 28. line 28 line reads long double y = log(1+(x));.

#include <stdio.h> #include <math.h> #define log(x) _generic((x),\   long double: log(1+(x))\ )  int main() {     double x;     printf("please enter number -1 +1: ");    scanf("%f", &x);     long double y = log(1+(x));     printf("from c math library: ln(1+x) = %f\n", y);  } 

if new programming, , trying make example work, not use macros @ stage focus on function , data type. start double typical working type of library interfaced math.h. example shows how calculate natural ln log() function.

as others have commented, scanf() needs format specifier of %lf double, printf() needs format specifier of %f double. note have tested return value scanf() function as as number input, check input behaved.

lastly, invite numbers in range -1 <= x <= 1 yet ln(0) undefined, hence have restricted input range exclude -1.

#include <stdio.h> #include <math.h>  int main(void) {     double x = 0, y;     printf("please enter number -1 +1: ");     if (1 != scanf("%lf", &x) || x <= -1 || x > 1) {         printf("i need -1 < number <= +1\n");         return 1;     }     y = log(x + 1);     printf("from c math library: ln(1+x) = %f\n", y);     return 0; } 

here sample outputs:

please enter number -1 +1: -1 need -1 < number <= +1  please enter number -1 +1: 0 c math library: ln(1+x) = 0.000000  please enter number -1 +1: 1 c math library: ln(1+x) = 0.693147