Question

需要编写一个函数，将字符串转换为double而不调用任何内置实用程序，例如atof()，Double.parseDouble()，Double.valueOf()，[NSString doubleValue]，{ {1}}，[NSScanner scanDouble]或任何类似的方法。

Answer 1

const char *s = "-123.456";
int sign = 1;
int hasdp = 0;
int dplaces = 0;
float f = 0.0;
while(*s) {
    switch (*s) {
    case '-': sign = -1; break;
    case '.': hasdp = 1; break;
    default:
        f *= 10;
        f += *s - '0';
        break;
    }
    s++;
    if (hasdp) dplaces++;
}

f *= sign;
int i;
for (i = 1; i < dplaces; i++) f /= 10;
// we need to start from 1 in order to get
// the correct result, this is a property
// of this algorithm

现在f将包含已解析的值，前提是其格式正确（可选 - 前缀，可选小数点后跟1个或多个小数位。

Answer 2

尝试以下方法：

- (void)myMethod:(NSString *)numberString{
    int i = 0;
    float number = 0.0; 
    while ([numberString length] > i) {
        NSString *myChar = [numberString substringWithRange:NSMakeRange(i, 1)];
        NSLog(@"%@", [numberString substringWithRange:NSMakeRange(i, 1)]);
        i++;
        if ([myChar isEqualToString:@"1"]){
            number = number * 10 + 1;
        } else if ([myChar isEqualToString:@"2"]){
            number = number * 10 + 2;
        } else if ([myChar isEqualToString:@"3"]){
            number = number * 10 + 3;
        } else if ([myChar isEqualToString:@"4"]){
            number = number * 10 + 4;
        } else if ([myChar isEqualToString:@"5"]){
            number = number * 10 + 5;
        } else if ([myChar isEqualToString:@"6"]){
            number = number * 10 + 6;
        } else if ([myChar isEqualToString:@"7"]){
            number = number * 10 + 7;
        } else if ([myChar isEqualToString:@"8"]){
            number = number * 10 + 8;
        } else if ([myChar isEqualToString:@"9"]){
            number = number * 10 + 9;
        } else if ([myChar isEqualToString:@"0"]){
            number = number * 10 + 0;
        }
    }
}

字符串到double而不调用任何内置实用程序

2 个答案: