这个怎么样-
NSString *formula = @"1+5*6";
NSExpression *exp = [NSExpression expressionWithFormat:formula];
NSNumber *resultForCustomFormula = [exp expressionValueWithObject:nil context:nil];
NSLog(@"%f", [resultForCustomFormula floatValue]);
编辑:
现在我考虑了您的要求并使用NSScanner 制作了一个方法您不会相信在Mr. Borrrden 建议我使用它之前我没有使用NSScanner,我发现它很棒。见下面的方法-
-(NSMutableString *)formatString:(NSString *)formula
{
// Let's check if there any wrong (.) value exm: 1/.2 or .7+3
// 1/0.2 and 0.7+3 are okay but above are incorrect so first fix them
NSString *str = formula;
NSInteger c = 0;
for(int i=0; i<[str length]; i++)
{
if([[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"+"] ||
[[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"-"] ||
[[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"/"] ||
[[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"*"])
{
if([str length] > i+1)
{
if([[NSString stringWithFormat:@"%c",[str characterAtIndex:i+1]] isEqualToString:@"."])
{
formula = [formula stringByReplacingCharactersInRange:NSMakeRange(i+1+c, 1) withString:@"0."];
c++;
}
}
}
}
// Now we will convert all numbers in float
NSString *aString;
float aFloat;
NSMutableString *formattedString = [[NSMutableString alloc]init];
NSScanner *theScanner = [NSScanner scannerWithString:formula];
while ([theScanner isAtEnd] == NO)
{
if([theScanner scanFloat:&aFloat])
{
[formattedString appendString:[NSString stringWithFormat:@"%f",aFloat]];
}
if([theScanner scanUpToCharactersFromSet:[NSCharacterSet decimalDigitCharacterSet] intoString:&aString])
{
[formattedString appendString:aString];
}
}
return formattedString;
}
这会将 (2.222/.4)+9999-7+0.7*.13 转换为 (2.222000/0.400000)+9999.000000-7.000000+0.700000*0.130000 .
在使用NSExpression之前调用这个方法。
NSString *formula = @"(2.222/.4)+9999-7+0.7*.13";
NSString *formattedString = [self formatString:formula];
NSExpression *exp = [NSExpression expressionWithFormat:formattedString];
NSNumber *resultForCustomFormula = [exp expressionValueWithObject:nil context:nil];
NSLog(@"Result = %f", [resultForCustomFormula floatValue]);
//OutPut: Result = 9997.646484
注意:我并不是说它适用于所有公式字符串。可能在某些情况下它不起作用。但它适用于一般方程。