strpbrk 不起作用

Question

char operators[] = "+-*/^(";  
    char* input = new char[100];  
    char* output = new char[100];  
    char* operadores = new char[100];  
    char* pch = input;  
    char* pch2 = input;  
    cout << "Expresion: " <<endl; cin.getline(input,100);  
    cout << input <<endl;  
    pch2 = strpbrk (pch2, operators);  
    pch = strtok (pch, "+-*/^(");  
    while (pch != NULL){  
        strcat (output, pch);  
        pch = strtok (NULL, "+-*/^(");  
        strcat (operadores, pch2);  
    }  

    cout << "Salida: " << output <<endl;
    cout << "Operadores: " << operadores <<endl;
    cout << "Entrada: " << input <<endl;
    cout << "pch2 = " << pch2 <<endl;

Hi, my problem is that the function strpbrk doesnt work, it doesn't return NULL. I've proved it, But I need a char to put in a stack. and cout doesn't show me what character is pointed by pch2.

Answer 1

You are confusing yourself - the program is designed to confuse.

You have both pch and pch2 pointing at the same input string - input . You call strpbrk() to find one of the operators, saving that position in pch2 . You then call strtok() on pch , and it finds the character that strpbrk() just found, and writes a NUL '\0' over it. So, it appears that pch2 points to the NUL at the end of a string. In the body of the loop, you then concatenate the empty string that pch2 points to onto your target list of operators.

Personally, I avoid using strtok() precisely because it mangles the input string. If you are going to use it, you are likely to need to work on a duplicate of the string, because strtok() writes NUL bytes over it.

Your diagnostic output at the end should show that the first section of the input - up to the first operator - only.

Beware of casting aspersions at standard library functions or compilers 'not working'. It is the mark of a tyro; 99.9999% of the time, it is a user error and not a system error. On those very, very, very rare occasions when you're correct (ooh, look - I've just won my third multi-million lottery prize; that's more likely, even though I don't buy lottery tickets), the way that you describe the problem is different. You describe the issue as one of utter astonishment; you document the working test cases; then you explain how the edge case you've found should work and the result - and you still aren't sure whether it is a bug in your code or in the system.

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As others diagnosed, you do not initialize operadores to an empty string, so concatenating to it leads to undefined behaviour.

There's really no need to allocate the 100-byte strings:

char input[100];  // Cleaner, simpler, more reliable