How to print the value at memory address?

[Shubhangi24]

I have written a C program without variable. And I want to print the value at that memory location.How to print that value?

code is like:-

int main()
 {
     printf("Enter value:");
     scanf("%d",1245024);
     /* how to print the value here */ 
 
 return 0;
 }

[donbock]

You want your program to read some arbitrary location that is outside the bounds of your program.

Doing so provokes implementation-defined behavior. Anything could happen, but the two most likely outcomes are a run-time error (typically a segment error) from trying to access an inaccessible memory location; or the memory location is successfully accessed.

Cast the address value into a pointer to the type corresponding to how you wish to access the location (that is, unsigned char* or unsigned short* or unsigned long* or ...). Then dereference the pointer and see what happens.

[Shubhangi24]

I want the answer without any variable not even pointer...

[whodgson]

Put the unsigned integer into the first element of an array and call printf on the array name. Also I don't think I have ever read (or heard)that a pointer could be classified as any form of variable.

[Banfa]

Shubhangi24: donbock did not suggest using a variable, he suggested that you cast you integer constant to have a pointer type and then dereferenced it to get the value at that address something like *((unsigned char*)1) no variables involved and a well known technique for accessing registers on an embedded platform..

whodgson: your array thing wont owrk and uses a variable. Also pointers can't be variables??? Think about it, what do you think have if you declare something like char *p; as used by countless C library functions. p is a variable and it is a pointer.

[whodgson]

Here's what the inventor says about it:
"The places in which we store data are called objects. To access an object we need a name. A named object is called a variable and has a specific type... the data items we put into variables are called values. The statement that defines a variable is (unsurprisingly) called a definition"

[whodgson]

to Banfa: I agree having tried it that an array won't even work for me presumably because a variable needs a name before memory is allocated and it becomes an object.
But I can't see how a pointer is also a variable because i understood that a valid pointer only ever contains the address of whatever it currently points at (or to). Although this has no bearing on the argument i also thought that a pointer had to be initialized at the time it was declared.

[donbock]

int a = 10;
int *b = &a;

a and b are variables.
The value of a is 10.
The value of b is the address of a. That is, b is a pointer to a.

[whodgson]

Yes...thanks, I've been back over it and what you both have said is what i have been reading. Some things just seem to go through to the keeper without registering.