Is it a pointer or a variable?

Question

Is it a pointer or a variable?

Debojit Acharjee 455

I want to know what is the difference between using the asterisk (*) before a pointer and after the "char" keyword? If "char a" means 'a' is pointer then while initializing it with a text why the asterisk () is not used before it? Does the compiler take it as a character variable or an integer variable containing t he memory address of the text?

Please see the following program:

#include <stdio.h>

char *food = "ICE CREAM";
char* drink = "COCA COLA";

int main ()
{

food = "BURGER";

    printf("%s, %s\n", food, drink);
    printf("%d %d", food, drink);

    return 0;
}

The output of this program shows the second initialized text "BURGER" and globally initialized text "COCOA COLA", when the pointer is converted to integer, it shows two integer values. What are these values?

YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-05-29T02:18:03.4+00:00
Hi Debojit Acharjee,

In C/C++, a pointer is a variable that stores a memory address. If you want to convert char* to int, you can refer to the following code:

const char* cPtr = "Hello"; int Value = reinterpret_cast<int>(cPtr);

You will see a numeric value that represents the pointer value. But this conversion may lead to undefined behavior, so it needs to be used with caution.
Debojit Acharjee 455 Reputation points

2023-05-29T04:50:58.95+00:00
This can be done pretty much easily using the "%x% format specifier and I have shown in the program below. I also have a question below.

#include <stdio.h> int main() { int *a; char *b; *a = 10; printf("The memory location of 10 is: %x\n", a); *a = 20; printf("The memory location of 20 is: %x\n", a); b = "C"; printf("The memory location of \"C\" is: %x\n", b); b = "Programming"; printf("The memory location of \"Programming\" is: %x\n", b); return 0; }

Output:

The memory location of 10 is: 2a8000

The memory location of 20 is: 2a8000

The memory location of "C" is: 4050aa

The memory location of "Programming" is: 4050cf

Every time I run this program, I see that the address of the integer pointers are always the same. Why is that?
Lamia Elshahat Eldesoky 76 Reputation points

2023-05-29T04:53:58.0533333+00:00

Good luck

Accepted answer

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Your answer

YujianYao-MSFT 4,296 Reputation points Microsoft External Staff

2023-05-29T02:18:03.4+00:00

Hi Debojit Acharjee,

In C/C++, a pointer is a variable that stores a memory address. If you want to convert char* to int, you can refer to the following code:

const char* cPtr = "Hello"; int Value = reinterpret_cast<int>(cPtr);

You will see a numeric value that represents the pointer value. But this conversion may lead to undefined behavior, so it needs to be used with caution.
Debojit Acharjee 455 Reputation points

2023-05-29T04:50:58.95+00:00

This can be done pretty much easily using the "%x% format specifier and I have shown in the program below. I also have a question below.

#include <stdio.h> int main() { int *a; char *b; *a = 10; printf("The memory location of 10 is: %x\n", a); *a = 20; printf("The memory location of 20 is: %x\n", a); b = "C"; printf("The memory location of \"C\" is: %x\n", b); b = "Programming"; printf("The memory location of \"Programming\" is: %x\n", b); return 0; }

Output:

The memory location of 10 is: 2a8000

The memory location of 20 is: 2a8000

The memory location of "C" is: 4050aa

The memory location of "Programming" is: 4050cf

Every time I run this program, I see that the address of the integer pointers are always the same. Why is that?
Lamia Elshahat Eldesoky 76 Reputation points

2023-05-29T04:53:58.0533333+00:00

Good luck

Answer 1

Bruce (SqlWork.com) 77,476 Volunteer Moderator

In C a pointer is an unsigned int. The sizeof(int) is the size of a pointer. Nowdays typically 32 or 64 bits. When you print a pointer as an int, you are printing the memory address it points to.

Also in C, the string literals datatype is char*.

Note: when printing a pointer you should use %i or large address may print as negative. Actually it’s common practice to print addresses as hex (%x).

Also C datatype declarations are one of its most difficult syntax’s to understand. The usage decorators are on the variable rather than the type (common on more modern languages)

// pointer to int array

int *myvar1[];

Debojit Acharjee 455 Reputation points

2023-05-29T04:08:58.6633333+00:00

Thank you, but what you mean by sizeof(int)? How come the token "int" would have a size? it's not a variable.

if char* is a type declaration for string laterals, then why an array is declared only using the "char" token (i.e. char array_name[size])?

Answer 2

Barry Schwarz 3,746

char a DOES NOT mean a is a pointer.

char *a means a is a pointer. The white space between the type (char), the asterisk, and

the variable name (a)is irrelevant. char*a, char* a, char *a, and char * a all mean the

exact same thing.

Since both food and drink are pointers, it is undefined behavior to print their values with %d.

You should be using %p. However, on most systems the undefined behavior produces the expected

results. In this case, the value printed for food is the address of the B in the string literal

"BURGER". Similarly, the value printed for drink is the address of the first C in "COCA COLA".

Finally, BURGER is not initialization text. Initialization can occur only when the variable is

being defined. food was originally initialized to point to "ICE CREAM". BURGER shows up in an

assignment statement that results in a new value being ASSIGNED to food.

RLWA32 49,301 Reputation points

2023-05-27T08:24:35.08+00:00
Expanding upon the information provided above by @Barry Schwarz note that for pointers the asterisk (*) binds to the variable name, not the type. Consider the following example where more than one variable is declared and initialized in a single line of code -

#include <stdio.h> int main() { char* p1 = "a", p2 = 'b'; // p2 is NOT a pointer char* p3 = "c", *p4 = "d"; //p3 and p4 are pointers char v1 = 'x', *v2 = "y"; // v2 is a pointer printf("p1: %p, p2: %c\n", p1, p2); printf("p3: %p, p4: %p\n", p3, p4); printf("p3 as 32-bit integer: %d\n", p3); // pointer value is truncated in 64-bit code printf("p3 as 64-bit integer: %lld", p3); return 0; }

Following image shows which variables are pointers while code is running.

Also, care must be taken when building 32-bit code versus 64-bit code if the code prints pointer values as integers. Note the difference in the printf format specifiers and the resultant output in which printing a 64-bit pointer as a 32-bit integer truncates the value and produces an incorrect result.
Debojit Acharjee 455 Reputation points

2023-05-28T06:42:40.64+00:00

Thank you for all your answers, but if the value of a pointer is displayed as a numeric address of the memory, is the pointer 'a' a constant integer type? Logically the declaration syntax of C has this format: type variable name = value (e.g. char a = 'H';), which means that the statement is telling the compiler that the variable 'a' can only store character value 'H', and oncourse, it will be stored in a memory with a numeric address. But if the statement is changed to char* a = 'H' would mean that 'a' should only store the address of the memory where 'H' will be stored. But I feel that here the type keyword "char" doesn't make sense because 'a' is not a variable but a pointer. So I think that the statement should be *a = 'H' and '*a' should be used everywhere in the program instead of only 'a' to retrieve the data from the memory location. This would be more meaningful to make a pointer standout from other variables.
Barry Schwarz 3,746 Reputation points

2023-05-28T14:01:11.8366667+00:00

The pointer is neither constant nor integer. The fact the everything in memory, including your variables, is stored in bytes that CAN be treated as integers has nothing to do with how C and C++ interpret the value of a variable.

The declaration char a = 'H' does not mean a can only store an 'H'. It means that when the variable a is created, it will be initialized with the value 'H'. That value can be changed by assignment and other techniques. It also means that a will be sized large enough to hold a single character, usually one byte.

The declaration char * a = 'H' does not mean that a should contain the address where the 'H' is stored in memory. It means that a should be sized large enough to hold the address of a character, usually 4 or 8 bytes, and should be initialized with whatever the value 'H' means when interpreted as an address. On a Windows system using ASCII, 'H' has the same value as 72. Thus a will contain the address 72 way down in low memory. For all practical purposes, the contents of memory location 72 is random.

Why do you think a pointer is any less a variable than an integer or a character?

*a used in a statement (as opposed to a definition) means that the system should dereference the pointer and use the value found at that address. For this to be well defined, it is essential that the pointer contain a value that points to memory you have access to. char * a = 'H' does not meat this condition. What you can do is

char c; reserve space for a character

char * a = &c; create the pointer and initialize it to point to c/

and sometime later execute * a = 'H'; assign the value 'H' to c, not to a

but someone reading your code would have to remember that a points to c, whereas

char c;

and sometime later c = 'H';

is both shorter and clearer.

But if you do have a pointer a that could point to either b or c or d at different times in your program, then *a would be the correct way to access the data that a is currently pointing to.
Debojit Acharjee 455 Reputation points

2023-05-29T04:29:06.0266667+00:00
I think you don't know the basics of C--you can't assign a single character value to a pointer using the single quotes. char *a = 'H' is not valid, but char *a = "H" is valid.

You didn't understand what I compared a pointer with variable. I didn't mean to say that a pointer is less than a variable, but it's expression should not be like a variable in a program, and the asterisk (*) should be there with the pointer where ever it is defined in a program. This will make easy for a programmer to read and make a difference between a variable and a pointer.

#include <stdio.h> int main() { char *a = "H"; a = "D"; printf("%s", a); return 0; }

In the above program, the pointer 'a' has been redefined with the character string 'D' and it was able to print it. Now my question is--if a didn't store the location of 'D' in the memory, then what did it store?
Barry Schwarz 3,746 Reputation points

2023-05-29T21:57:45.2666667+00:00

The pointer a was NEVER redefined at all. Your assignment statement simply changed the value of the pointer, that is its contents. This is no different than int b = 1 followed later by b = 2. You are allowed to changed the value that is stored in a variable. That is why they are called variables and not constants.

Are you aware that both "H" and "D" are each an array of two char? Your initialization caused a to contain the address of the first char in the first array. Your assignment statement replaced that value with the address of the first char in the second array, just as the 2 replaced the 1 in the integer b described above.

Answer 3

Lamia Elshahat Eldesoky 76

The same answer

Debojit Acharjee 455 Reputation points

2023-05-29T05:03:50.2+00:00
I don't know which same answer you are talking about, but can you answer this one:

Why the following program can't store the number '7' of the floating point number "12345678901234567", but displays '8' instead of '7' of the last digit?

#include <stdio.h> int main() { double a; a = 12345678901234567; printf("%lf", a); return 0; }

Output:

12345678901234568.000000

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Is it a pointer or a variable?

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