Compose a prgrm that directly displays the byte-sizes of the standard data types in C

[gauravrishi]

hello guys...
pls help me in this....
i m new to this language...C

help me !

[gauravrishi]

hello guys...
pls help me in this....
i m new to this language...C

Compose a program that directly displays the byte-sizes of the standard data types in C including:
char, short, int, unsigned int, long, unsigned long, float, double, long double, and void. Use the
sizeof() C function that accepts any data type or identifier as a parameter and returns an integer
describing the number of bytes used.
Sample statement showing the size of an int,
printf("The amount of memory used by an int is: %d bytes\n", sizeof(int));

help me !

[gauravrishi]

1. Compose a program that directly displays the byte-sizes of the standard data types in C including:
char, short, int, unsigned int, long, unsigned long, float, double, long double, and void. Use the
sizeof() C function that accepts any data type or identifier as a parameter and returns an integer
describing the number of bytes used.
Sample statement showing the size of an int,
printf("The amount of memory used by an int is: %d bytes\n", sizeof(int));
2. Add to the above program statements to display:
– The size of an array and an array pointer.
Declare an array of 4 integer elements called myArray[ ]
Display sizeof(myArray[4]) – this displays the size of the array
and sizeof(myArray) – this displays the size of the array pointer
Question: Does the program know how "big" the array is?
– The size of an explicit pointer.
Declare in integer pointer (int *ptr)
Display sizeof(ptr) and sizeof(int *)
Question: Does the size of the pointer differ if the pointer is changed to a double ?

With the following code, don't forget to use correct prototyping when writing functions!
3. Factorial
Write a function called factorial().
This function accepts one int parameter (n) and returns an int value which is the result of calculating
n!. The value of 5-factorial or 5! = 1 * 2 * 3 * 4 * 5.
The factorial calculation must be done using a simple for-loop, do not use recursion.
Test your function with a call from main(). Use a few different values, along with 0! and 1! (they
should both = 1).
4. Sum the Digits
Write a function called sumDigits().
This function accepts one int parameter (n) and returns an int value which is the sum of the digits
that makeup n. A call to sumDigits(512) will return the value of 5 + 1 + 2. The function must
work with all values of n where n >= 0, if n < 0, sumDigits() returns 0.
Test your function with a call from main() using a few different values.
5. Sum the Array
Write the function sumArray().
This function accepts an array filled with doubles as a parameter as well as an int which is the
number of elements in the array. The function returns a double value which is the sum of all of
the values in the array.
Test your function with a call from main() using at least 3 different sized arrays each containing
different values (the same value should not be repeated in any of the arrays).

[gerard4143]

Try using(including) the limits.h header file into your program.

[Ancient Dragon]

There is nothing in limits.h that will help in this assignment. You will need to use the sizeof operator to obtain byte sizes of data.

[Narue]

We're not here to do your homework for you. Please show that you've put some effort into doing these exercises on your own.

[cse.avinash]

sizeof() is used to calculate the sizes of datatypes, in number of bytes.

It returns the size of the type of the variable or parenthesized type-specifier that it precedes as a size_t type value. sizeof can be applied to all datatypes, be they primitive types such as the integer and floating-point types defined in the language, pointers to memory addresses, or the compound datatypes (unions, structs, or C++ classes).

for eg.

printf("%d",sizeof(char));

will return the size occupied by int in memory i.e., 1 byte.
similarly you can do for others .
Do the coding part and if still there is a problem , we are there for you to help.