字符串转换为整数而不是双精度

Question

I want my function to convert a string to a double but if I type for example '22.4' it will output '22'. How do I get it to output a double?(USING fgets) I've read other questions that had answers saying use sscanf but is there a way for using fgets?

double getDouble(char message[]){

double val;
double var = 0.0;
char input[80];

do{
    printf("%s\n", message);
    fgets(input, 80, stdin);

    if (input[strlen(input)-1] == '\n'){
        input[strlen(input)-1] = '\0';
    }
    val = atof(input);

    if (val != 0.0){
        var = 1.0;
    }
    else{
        printf("Error. Please try again.(Only entering numbers is valid)\n");
    }
}while(var == 0.0);
printf("The result is: %f\n", val);

return val;


}

Answer 1

Here is a working example.

Some modifications:

• Using a near_equal() function, as testing equality of doubles is dangerous.
• Instead of using a double flag such as var = 0.0 , why not just use an int ?

• Checking the return value of fgets is always a good idea.

   #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #define BUFFSIZE 80 #define EPSILON 1e-8 double getDouble(char message[]); int double_equal(double a, double b); int main(void) { double num; num = getDouble("Enter a real number: "); printf("The result is: %.1f\\n", num); printf("The number you entered was %g\\n", num); return 0; } double getDouble(char message[]) { double val; char input[BUFFSIZE]; int found = 0; do { printf("%s\\n", message); if (fgets(input, BUFFSIZE, stdin) != NULL) { if (input[strlen(input)-1] == '\\n'){ input[strlen(input)-1] = '\\0'; } val = atof(input); if (!double_equal(val, 0.0)) { found = 1; } else { printf("Error. Please try again.(Only entering numbers is valid)\\n"); } } } while (found == 0); return val; } int double_equal(double a, double b) { return fabs(ab) < EPSILON; }