CODEWITHSUNDEEP
c++
Zack Oliver
Zack Oliver

Reputation: 107

Why can't I compare char to "*"?

Why can't I compare UserInput[i] to "*" ? The compiler says "ISO c++ forbids comparison between pointer and integer". Where is the integer? The only one I see on that line is i, which is used for locating the specific character.

#include <iostream>
#include <fstream>
#include <string.h>

using namespace std;

int main(){
    char UserInput[4] = "2*3";
    for (int i=0; i< strlen(UserInput); i++)
    {
        if(UserInput[i] == "*")
        {
            cout<<"There is multiplication Involved \n";
        }
    }
    return 0;
}

Upvotes: 1

Views: 861

Answers (4)

fat
fat

Reputation: 7123

It's a little bit tricky.
All inside " " is a string - array of chars. Array in c/c++ is pointer to first element of array.
Each single element of string is char, which is converted to int.
At the end, compiler says that it's not possible to compare this two items. You need to do like this:

if(UserInput[i] == '*')

In this case you compare two chars with no problem.

PS If you want to determine presence of one string in another string, you should use strstr().

Upvotes: 0

Hariom Singh
Hariom Singh

Reputation: 3632

if(UserInput[i] == "*")

Will issue a warning because of ""

"*" is not char. "*" is a const char*.

Comparison between pointer and integer (int and const char *) is not possible

change it to

if(UserInput[i] == '*')

Upvotes: 2

Shreevardhan
Shreevardhan

Reputation: 12641

Additionally, use std::string instead of char arrays

#include <iostream>
#include <string>
using namespace std;

int main() {
    string UserInput = "2*3";
    for (int i = 0; i < UserInput.length(); i++) {
        if (UserInput[i] == '*') {
            cout << "There is multiplication Involved \n";
        }
    }
    return 0;
}

Upvotes: 1

R Sahu
R Sahu

Reputation: 206667

Where is the integer?

Your conditional expression is UserInput[i] == "*".

In the expression UserInput[i] is of type char. In many of the binary operations, char is promoted to int before the operation is carried out. As far as the compiler is concerned, it is comparing an int with "*", which is of type const char[2] but decays to a pointer in the expression.

What you need to do is compare the char with the character constant '*'. Use 

if(UserInput[i] == '*') {

Upvotes: 5

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