Difference in behavior between a char* and an int* in C++

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Have two snippets below . Need to understand why char* is successful while int* gives an error. Trying to understand the difference in behavior between a char* and an int*
Thanks in advance for your help .

What I have tried:

snippet 1 :

int *p;
p=19;

cout<<"p val"<<*p<<endl;
cout<<"p addr"<<&p<<endl;

o/p:
-----
main.cpp: In function ‘int main()’:
main.cpp:14:3: error: invalid conversion from ‘int’ to ‘int*’ [-fpermissive]
p=19;
^~
===============================================================

snippet 2:

char *a;
a="fubar";

cout <<"a val = "<<*a<<endl;
cout <<"a = "<<a<<endl;
cout <<"a addr = "<<&a<<endl;

o/p:
-----

$g++ -o main *.cpp
main.cpp: In function ‘int main()’:
main.cpp:14:8: warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
a="fubar";
^~~~~~~
$main
a val = f
a whole = fubar
a addr = 0x7ffed3404898
===============================================================

Posted 17-Sep-18 10:06am

out0FBox

Updated 17-Sep-18 10:58am

v2

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4 solutions

Solution 1

The message says it all - you can't convert a string constant to a 'char *'. The reason being that it would then possible to change the constant and that is not allowed. To make this compile and run, change the declaration of a to 'const char *a' and then it will work.

-edit-

Regarding snippet 1 : it is declared as a pointer and the next line assigns the value 19 to that pointer. That is not likely to be a valid address and that is what the message is referring to. Also, you probably don't really want the address of the pointer but its value which would be the address it holds. This means that line should be changed :

cout << "p addr" << p << endl;

and that will display the address the pointer holds. Here is a better use of a pointer to an integer :

C++

int ivalue = 0;
int* iptr = &ivalue;   // set iptr to the address of ivalue
*iptr = 42;            // set the value of the integer pointed to

now ivalue will have a value of 42. The key thing here is the value of a pointer is the address it holds. To get the data at that address you have to dereference the pointer and that is what the asterisk does.

Posted 17-Sep-18 10:21am

Rick York

Updated 18-Sep-18 6:28am

v3

Comments

out0FBox 17-Sep-18 16:27pm

Hi Rick,
Thanks for the reply.
However, the warning for char* is not the issue . I wanted to know why it failed for snippet 1 .

Regards

Solution 3

A constant character string in C++ is of type 'const char*', which most compilers will allow you to assign to a 'char*' (possibly with a warning). OTOH, an 'int' (e.g. 19) is never compatible with 'int*'. You must dereference the pointer thus:

*p = 19;

(There is a minor exception to this rule - a pointer may be assigned the value 0, which in this context is equivalent to nullptr. However new programs should use nullptr rather than 0.)

Posted 17-Sep-18 10:29am

Daniel Pfeffer

Updated 17-Sep-18 10:32am

v3

Comments

out0FBox 17-Sep-18 16:40pm

Thanks for the immediate response . I was looking for an answer close to what OriginalGriff has posted above. But this helps adds too.

Regards

Solution 4

The comparison is not fair: code snippets are not equivalent.
You have to compare either

C++

// snippet a.1
int *p;
p = 19;

cout << "p val"<< *p << endl;
cout << "p addr" << p << endl;


// snippet a.2
char *s;
s = 'a';

cout << "s val"<< *s << endl;
cout << "s addr" << s << endl;

Compiler output:

[...]: error: invalid conversion from ‘int’ to ‘int*’ [-fpermissive]
  p = 19;
      ^~
[...]. error: invalid conversion from ‘char’ to ‘char*’ [-fpermissive]
  s = 'a';
      ^~~

or

C++

// snippet b.1
const int ia[] = {1,2,3};
int *p;
p = ia;

cout << "p val"<< *p << endl;
cout << "p addr" << p << endl;

// shippet b.2
char *s;
s = "alpha";

cout << "s val"<< *s << endl;
cout << "s addr" << s << endl;

Compiler output

[...]: error: invalid conversion from ‘const int*’ to ‘int*’ [-fpermissive]
  p = ia;
      ^~
[...]. warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
  s = "alpha";
      ^~~~~~~

In the above case (b snippets), the compiler actually consider char somehow special (I guess it is a C legacy).

[update]

Quote:
But what i do not know is why a regular integer pointer cannot point to a const integer.

A pointer can point to an integer constant, for instance:

C

const int myconst = 10;
const int * p = &myconst; // p is a pointer to const (NOT a const pointer)
printf("Address = %p, value = %d\n", p, *p);

However you cannot arbitrarly assign an address to it (as in your example):

const int * p = 19; // ERROR

is plainly wrong.

Quote:
My first assumption is that , allowing so would allow the value of the const int to be changed through the pointer

This is fairly correct: you can assign the address of a integer const only to a pointer to integer const (NOT a const pointer).

Quote:
econd assumption is that a const pointer is to be used to point to a const variable

Nope.

A pointer to integer const is to be used to point a integer const.
A const pointer is a pointer whoose value (that is the address it points) cannot be changed.

E.g.

C

const int myconst = 10;
  const int * p = &myconst; // p is pointer to const int
  int i = 0;
  int j = 0;
  int * const q = &i; // q is a const pointer to int
  *q = 50; // i.e. i=50, it is OK, since q points to a variable 
   q = &j; // ERROR, cannot change the address q points to

[/update]

Posted 17-Sep-18 10:58am

CPallini

Updated 19-Sep-18 22:18pm

v2

Comments

out0FBox 19-Sep-18 18:51pm

Yes. The comparison is off in my example question.
As per snippet b.1, if the array is not a cont integer, there should be no error.
But what i do not know is why a regular integer pointer cannot point to a const integer.
My first assumption is that , allowing so would allow the value of the const int to be changed through the pointer. Second assumption is that a const pointer is to be used to point to a const variable.
Do let me know if my assumptions are correct.

CPallini 20-Sep-18 4:18am

See my updated solution.

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OriginalGriff · Accepted Answer · 2018-09-17T10:29:00

C++

int *p;
p=19;

p is a "pointer to an integer", and 19 is an integer, not an address, so the compiler is complaining that you can't assign an integer directly to a pointer.

C++

char *a;
a="fubar";

a is a "pointer to a character", but "fubar" is a string, which a a null-terminated array of characters. And the name of an array is a pointer to the first element of the array - that is in the language definition, and is fundamental to how the language works.
Since "fubar" is made of characters, it is effectively (and really) a "pointer to the first character in the string" so you can assign it without problems.

Difference in behavior between a char* and an int* in C++

4 solutions

Solution 1

Solution 2

Solution 3

Solution 4

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