How to assign a String to a variable in C

Question

I create an Entity object:

#ifndef ENTITY_H  
#define ENTITY_H  

struct MyEntityObject {  
    char  _entityAuthor;
};  

#endif // ENTITY_H

Then I try to set it up and use it:

struct MyEntityObject myEntityObject;  

/* MyEntityObject data specification */  
strcpy(myEntityObject . _entityAuthor, "Shakespear");

I get this error:

main.cpp:37: error: invalid conversion from ‘const char*’ to ‘char’ [-fpermissive]  
      myEntityObject . _entityAuthor = "Shakespear";  
                             ^

What is going on here? What am I getting wrong?

Stephen Docy · Accepted Answer · 2018-05-01 00:39:26Z

2

Your struct field is only a single char

struct MyEntityObject {  
    char  _entityAuthor;  
};

Trying changing it to either a char * that you dynamically allocate memory to or a char array if you can make solid assumptions about the size of the string.

answered May 1, 2018 at 0:39

Stephen Docy

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1 Comment

bruceg Over a year ago

Just to clarify the meaning of dynamically allocate above, if the OP uses a char * for the field, then the call to strcpy will crash. strdup will allocate the memory for the field, if malloc hasn't been called.

Ronan Boiteau · Accepted Answer · 2018-05-01 00:47:35Z

0

This line:

char _entityAuthor;

...declares a char variable.

What you need is a pointer to char (char *) variable:

char *_entityAuthor;

You will also need to allocate memory with malloc() before using strcpy().

Here is a tutorial.

edited May 1, 2018 at 0:47

answered May 1, 2018 at 0:40

Ronan Boiteau

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