C Data Types, Variables & Storage Classes

The initial address of char and float data types is 0.
So, the values of (int)(pc+1) and (int)(pf+1) will be 1 and 4, respectively.
Hence, the output of the code will be 5.

More than one global variable declaration is allowed. But you can initialise only one variable. Hence, output will be 8.

The value of ch is outside the range of char (-128 to +127).
So, it goes to other the side and becomes 65.
Hence, the ASCII value of 'A' = 65

The register variable stores in Central Processing Unit (CPU).
So, y does not have any memory location.
Hence, py=&y is not a valid statement.

Rch is a character variable and it occupies 1 byte.
Rn is an integer variable and it occupies 2 bytes.
Rr is a float variable and it occupies 4 bytes.
Hence, the output will be 1 2 4. 

First printf() will print 7, then if condition gets true due to non-zero integer and t will decrement and main() call recursively.
Again printf() will print 6, then if condition gets true due to non zero integer and t will decrement and main() call recursively and so on.
If condition gets false, when value of t will be 0.
Hence, output will be 76543210. (The static variable always retains its value)

The ASCII value of 'T' is 84.
So, the value of T + 6 is 90.
Hence, '%c' converts the integer 90 to character 'Z'.

Here storage class static is not allowed. 

The range of int is - 32768 to + 32767.
40000 is exceeding + 32767 by 7233 and 45000 exceeding + 32767 by 12233.
So, 7233^th and 12233^th integers from the other side of range is - 25536 and - 20536.

Here 't' is a pointer to a constant integer. But program wants to change the value of t by statement ++(*t), which is not possible. Hence, it will give the error 'Cannot modify the constant object'.

Here '/**/' is the comment and is the simple statement like d = t/c.
Hence, it gives the output 5.

Res = (t+c)/8 = (9.0 + 78.50)/8 = 87.50/8 = 10.937500
Due to '%.4f', it will print 10.9375.

'j' is used before the declaration.
Hence, the compiler will give the 'Undefined symbol j' error.

For sinx, cosx, e^x and log₁₀x , the functions are sin(x), cos(x), exp(x) and log10(x), respectively.
Hence, the correct expression is pow(sinx,2) + cos (2*x) + sin(exp(x) + log10(x)).

Factorial of 8 is out of range from the int data type (-32768 to +32767).
Factorial of 8 is 40320, i.e. it goes to other end and displays the (40320 - 32767) = 7553^th term from the -ve side.
Hence, the output will be - 25216.

The sum from 0 to 922 is 32287.
If 923 is added with 32287, the sum will be 33210, which is outside the range of integer data type.
Hence, n up to 922 can give the desired output.

It runs properly if the message is to be typed as it is in the scanf() statement, and then enter the values of t and c.
For example,
After running the program:
Enter the values of t and c = 45 67
The output will be t = 45 c = 67

The range of long signed int is - 2147483648 to + 2147483647.

In C int, datatype is of 2 bytes while char pointer pt will take only one byte.
Binary representation of 32 is 00000000 00100000. 
The pr pointer will first point the second byte, and then the first byte.
So, second byte will be represented by 00100000, i.e. 32 and first byte will be represented by 00000000, i.e. 0.
Hence, the output will be 32 0.

Loop will execute for i = 65 to 71.
The ASCII value of 'A' = 65
and ASCII value of 'G' = 71
Hence, the output will be “ABCDEFG”.

Here, 't' is the external variable and specifies to the compiler that the memory for 't' is allocated in some other program. Hence, linker error has occurred.

If any signed integer data type takes n bytes, the range of that data type will be -2^8n-1 to +2^8n-1 - 1.
In this case, n = 4. 
The maximum value stored = +2^{8 x 4 - 1} - 1 = + 2³¹ - 1