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Home/ Questions/Q 9247107
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T09:39:00+00:00 2026-06-18T09:39:00+00:00

I’m new to c++, I’m writing a program that reads code from a file,

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I’m new to c++, I’m writing a program that reads code from a file, and classifies each part of it as an identifier, bracket, keyword, etc..

I’m doing this using if else statements, it works fine except with brackets and semicolons.

for example if(a== "=" ) cout << "a is the equal operator" works, but if(a== ";" ) cout << "a is a semicolon" doesn’t. I also tried using the compare method, it didn’t work either.

Can someone please tell me why that’s happening?

Thanks

void checkString(string a)

        if(a=="("){
                cout << "RPAR: " << a + "\n";
        }

        else if(a==")"){
                cout << "LPAR: " << a + "\n";
        }

        else if(a.compare("{") == 0){
                cout << "LBRAC: " << a + "\n";
        }

        else if(a=="}"){
                cout << "RBRAC: " << a + "\n";
        }

        else{
                cout << "IDENTIFIER: " << a + "\n";
        }
}

int  main (){

    std::vector<string> STRINGS;
    string STRING;
    ifstream infile;
    infile.open("m.c");
    while(getline(infile,STRING,' ')){
            STRINGS.push_back(STRING);
    }

    infile.close();
    for(int i=0; i<STRINGS.size(); i++){
            checkString(STRINGS[i]);
    }

    return 0;
}

If a is a bracket or a semicolon the program prints IDENTIFIER: {.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T09:39:02+00:00Added an answer on June 18, 2026 at 9:39 am

    It looks like the problem is with getline(infile,STRING,' ').

    What happens if your code hits a \n, \t, etc? It will be extracted together with your token and you will end up with a string such as ";\n", which is not the same as ";"

    Either change the logic that you use to tokenize your file:

    std::ifstream fin("m.c");
    while (fin >> STRING ){
       STRINGS.push_back(STRING);
    }
    fin.close();
    

    Or trim your string.


    Update:

    And keep in mind that this kind of tokenizer will only work if your tokens are separated by whitespaces (return 0 ; will work as expected, return 0; will be tokenized to return and 0;)

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