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问题:
It tells me that it can't convert int to bool.
Tried TryParse but for some reason the argument list is invalid.
Code:
private void SetNumber(string n)
{
// if user input is a number then
if (int.Parse(n))
{
// if user input is negative
if (h < 0)
{
// assign absolute version of user input
number = Math.Abs(n);
}
else
{
// else assign user input
number = n;
}
}
else
{
number = 0; // if user input is not an int then set number to 0
}
}
回答1:
You were probably very close using TryParse
, but I'm guessing you forgot the out
keyword on the parameter:
int value;
if (int.TryParse(n, out value))
{
}
回答2:
Just use this:
int i;
bool success = int.TryParse(n, out i);
if the parse was successful, success
is true
.
If that case i
contain the number.
You probably got the out
argument modifier wrong before. It has the out
modifier to indicate that it is a value that gets initialized within the method called.
回答3:
private void SetNumber(string n)
{
int nVal = 0;
if (int.TryParse(n, out nVal))
{
if (nVal < 0)
number = Math.Abs(nVal);
else
number = nVal;
}
else
number = 0;
}
回答4:
You can try with some simple regular expression :
bool IsNumber(string text)
{
Regex regex = new Regex(@"^[-+]?[0-9]*\.?[0-9]+$");
return regex.IsMatch(text);
}
回答5:
You could try something like below using int.TryParse
..
private void SetNumber(string n)
{
int parsed = -1;
if (int.TryParse(n, out parsed)) //if user input is a number then
...
The reason there are complaints that it cannot convert an int
to a bool
is because the return type of int.Parse()
is an int
and not a bool
and in c# conditionals need to evaluate bool
values.
回答6:
int.Parse will give you back an integer rather than a boolean.
You could use int.TryParse as you suggested.
int parsedValue;
if(int.TryParse(n, out parsedValue))
{
}
回答7:
There are a lot of problems with this code:
- Using VB-style line comments (') instead of C# slashes
- Parse for integer returns an int and not a bool
- You should use TryParse with an out value
- h does not seem to be valid at all. Is it a type for n?
- There are variables that do not seem to be defined in function scope (number) are they defined at class scope?
But try this:
private void SetNumber(string n)
{
int myInt;
if (int.TryParse(n, out myInt)) //if user input is a number then
{
if (myInt < 0) //if user input is negative
number = Math.Abs(n); //assign absolute version of user input
else //else assign user input
number = n;
}
else number = 0; //if user input is not an int then set number to 0
}
回答8:
int.Parse will convert a string to an integer. Current you have it within an if statement, so its treating the returned value of int.Parse as a bool, which its not.
Something like this will work:
private void SetNumber(string n)
{
int num = 0;
try{
num = int.Parse(n);
number = Math.Abs(num);
}catch(Exception e){
number = 0;
}
}
回答9:
Well for one thing the inner if statement has an 'h' instead of an 'n' if(h < 0). But TryParse should work there assuming that 'number' is a class variable.
private void SetNumber(string n)
{
int temp;
bool success = Int32.TryParse(n, out temp);
// If conversion successful
if (success)
{
// If user input is negative
if (temp < 0)
number = Math.Abs(temp); // Assign absolute version of user input
else // Assign user input
number = temp;
}
else
{
number = 0;
}
}
回答10:
//vinojash@gmail.com
//In my knowledge i did this in simple way thanks for watch my code
static void Main(string[] args)
{
string a, b;
int f1, f2, x, y;
Console.WriteLine("Enter two inputs");
a = Convert.ToString(Console.ReadLine());
b = Console.ReadLine();
f1 = find(a);
f2 = find(b);
if (f1 == 0 && f2 == 0)
{
x = Convert.ToInt32(a);
y = Convert.ToInt32(b);
Console.WriteLine("Two inputs r number \n so tha additon of these text box is= " + (x + y).ToString());
}
else
Console.WriteLine("One or tho inputs r string \n so tha concadination of these text box is = " + (a + b));
Console.ReadKey();
}
static int find(string s)
{
string s1 = "";
int f;
for (int i = 0; i < s.Length; i++)
for (int j = 0; j <= 9; j++)
{
string c = j.ToString();
if (c[0] == s[i])
{
s1 += c[0];
}
}
if (s==s1)
f= 0;
else
f= 1;
return f;
}