# V731\. The variable of char type is compared with pointer to string\.

The analyzer has detected a comparison of a char variable with a pointer to a string\. The reason why the variable is used that way is in using double quotes \("\) instead of single quotes \('\) by mistake\. 

Here's an example for this error pattern:

```cpp
char ch = 'd';
....
if(ch == "\n")
....
```

The inattentive author of this code wanted to compare the 'ch' variable with a new string's character but used quotes of a wrong type\. This resulted in the value of the 'ch' variable being compared to the "\\n" string's address\. Code like that can compile and execute well in C but usually makes no sense\. The correct version of the code sample above should use single quotes instead of double ones:

```cpp
char ch = 'd';
....
if(ch == '\n')
  ....
```

The same kind of mistake can be also made when initializing or assigning a value to a variable, causing this variable to store the least significant byte of the address of the string being assigned\.

```cpp
char ch = "d";
```

The correct version of the code should use single quotes\.

```cpp
char ch = 'd';
```