sizeof Char / Array length Memory allocation in C,

Question

Can somebody tell me about the memory allocation in c? What is the size of a char variable? 2 or 4? Why the difference in the address value between 2 neighboring char elements in an array is only 1?

char foo [] = {'a', 'b'};
printf ("This is the Address of val1 %d \n", &foo[1]);           // -1079295441
printf ("This is the Address of val2 %d \n", &foo[2]);           // -1079295440
printf ("The size of each array member is %d \n", sizeof(foo));  // 2

foo is a character array and as such has the same size as any other array since it's essentially just a pointer. — apokryfos
– apokryfos, Commented Sep 26, 2016 at 0:07
@apokryfos, arrays are not pointers. They have various sizes depending on their element type and capacity. — John Bollinger
– John Bollinger, Commented Sep 26, 2016 at 0:30

Sergey Kalinichenko · Accepted Answer · 2016-09-26 00:07:29Z

3

You are not printing addresses correctly: an address is not an int, so you cannot use %d to print it. Use %p (for "pointer") instead, and cast the address to void* for printing:

printf ("This is the Address of val1 %p\n", (void*)&foo[1]);

Now your program produces this or similar output:

This is the Address of val1 0xffbd2fcf 
This is the Address of val2 0xffbd2fd0 
The size of each array member is 2

Two pointers are off by 1, which is the size of a single char. The size of the entire array is 2. If you want to print the size of a single element, use foo[0].

answered Sep 26, 2016 at 0:07

Sergey Kalinichenko

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4 Comments

M.M Over a year ago

@Oka There's no excessive cast in this answer

Sergey Kalinichenko Over a year ago

@Oka It's required. %p wants void*, so passing int* is considered UB.

Oka Over a year ago

Ah, this is an exception to the safe promotion of any pointer to void *. Guessing this has to do with varargs? Interesting.

Sergey Kalinichenko Over a year ago

@Oka Although any pointer can be used in the context where void* is needed (a call to a function with void* parameters, an assignment to a void* variables) printf has a variable-length argument list. Hence, by default pointers go unconverted.

mpekim · Accepted Answer · 2020-07-22 13:33:06Z

Memory Allocation is a bit tricky, but it's easier than you think. It is, as the name implies, the usage and selection of data to optimize program function and speed. Basically, Memory Allocation is a "perk" of languages such as C/C++ that allow programmers to only use EXACTLY as much data as needed, freeing up memory for other computer functions.

Some good info to know about memory...

memory is known in "bytes", these are 8-bit "groups" of data.
A "bit" is a value that is either 0 or 1.

Variable sizes:

Char: 1 byte
Int: 4 bytes
double: 8 bytes

When neighboring elements are of the same type (for example, an array of chars), they will have a difference in address that increments/decrements by the value of the memory size. Since a char variable has a size of 1 byte, neighboring elements will have addresses that differ by 1.

Ex: char addresses: 1204, 1205, 1206, 1207... (1 byte)

int addresses: 1204, 1208, 1212, 1216... (4 bytes)

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