C ++字符串和char的比较

Question

I am trying to compare a string and a char like below :

char filter = "    " // <<== this is a tab space, so I am assuming its one char
if ( line[counter] == filter[0]){

}

note that line is a normal string defined like : string line; . now for some reason the statement is never true even though there are no syntax errors.

UPDATE

inside line is : string line = "1 90 74 84 48 76 76 80 85";

2nd UPDATE

here is the complete function I wrote :

void getResults(string line){

    int tmpSize = line.length();
    int counter = 0;
    int tmpCounter = 0;

    while(counter != MAX_No_Of_Grades){

        if(line[counter] == '\t\t'){
            counter++;
        }else{
            cout << tmpCounter << ". this is : " << line[tmpCounter] << "a" << endl;
            tempGrade[counter] += line[tmpCounter];
        }
        tmpCounter++;
    }


}

the function is technically suppose to break the string line into an array by "tabspace". but right now counter does not change and therefore its an endless loop!

Answer 1

I'm not sure what code you are actually running, because the char filter = " " line has no terminator and that assignment is definitely illegal. You cannot assign the string literal (a char array) to a single char variable.

If you want to determine whether the nth character of a string is a tabulation, the following code would probably be what you are looking for:

if (line[counter] == '\t') {
  // match...
}

---

As for updates. If typed as string line = "1 90 74 84 48 76 76 80 85"; in your program, there are no tabulations in this string. There are only spaces. Moreover, '\\t\\t' is a pair of tables. Put a single \\t in there for a single tab.

Here is a slightly modified version of your sample function:

void getResults(string line)
{
    int tmpSize = line.length();
    int counter = 0;
    int tmpCounter = 0;
    while((tmpCounter < tmpSize) && (counter != MAX_No_Of_Grades))
    {
        if(line[counter] == ' ') {
            counter++;
        }
        else {
            cout << tmpCounter << ". this is : " << line[tmpCounter] << "a" << endl;
            tempGrade[counter] += line[tmpCounter];
        }
        tmpCounter++;
    }
}

This version has an extra check in the while loop to stop consuming characters after the end of line is reached. It also uses a space character since your test input does not use tabulations.

---

If this is not homework and you can use all the standard library facilities you want, I would suggest looking into more advanced input strategies. Here is a simplified version of your function to extract every single number in the array.

#include <iostream>
#include <string>
#include <sstream>
void getResults ( const std::string& line )
{
    std::istringstream input(line);
    for (int grade=0; input >> grade;) {
        // process grade.
    }
}

If you want to remove the global variable and handle any number of grades, you can use a std::vector<> to automatically increase the "array" size as you get more and more grades.

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
std::vector<int> getResults ( const std::string& line )
{
    std::istringstream input(line);
    std::vector<int> grades;
    for (int grade=0; input >> grade;) {
        grades.push_back(grade);
    }
    return grades;
}

Answer 2

char filter = "    " // <<== this is a tab space, so I am assuming its one char

What you are assuming is wrong. A tab space char is '\\t' .

What you have there instead is multiple space characters.

Edit:

Also this : char filter = " " is wrong

This const char * filter = " "; //is more like it...

But this : char = '\\t'; is maybe what you wanted.

Answer 3

Ok. Lets go over you code by parts to make things clear.

char filter = "    " // <<== this is a tab space, so I am assuming its one char

Your first statement creates a variable of type char. The type char in c++ take a value, like any other variable. The only difference is that if you try to use it with a statement, like cout, it will be interpreted as a character by it's ASCII value. What you did compiles and works, but the char you declared has the value that c++ gives for the string " ". The value for the horizontal tab (TAB) in ASCII is 9.

So, for declaring a char of datatype char with the value for a TAB:

char filter = 9;

But you don't have to keep remembering all those values. The same way C++ will give a value for the string " ", it will give a value for a character. If you want to get the value for characters, you have to use ''. A confusion that some make is to assume that tab is actually a group of spaces. Tabs are interpreted, by a text editor, as a group of spaces. But it is in fact a single character. In C++, the character for the tab is \\t. If you actually put this inside a string and call cout, you will have a tab. The reason why \\t is a single character and not two characters is because the compiler gives "special" interpretations for the immediate character after a backslash inside '' or "".

So the following will give the value of 9 to a variable of datatype char:

char filter = '\t';

Now for your if statement:

if ( line[counter] == filter[0])

Your if statement is comparing the character with index "counter" in the string line. This is possible because a string is an array of characters (It is actually a class, but don't worry about it for now). But filter is not an array, it is a single character. So it wont have an index (Eg [0]). All you need is the variable you created. This confusion usually happens because the concept of string is often introduced without explaining that a string is not a datatype (That's why is not blue, and that's why you have to include a library).

So, just removing the index you will have:

if (line[counter] == filter)

You could, of course, just compare it to a tab directly

if (line[counter] == 9);

or

if (line[counter] == '\t');

Hope this gives you a better understanding of what is actually happening.