I.11: What is an rvalue?

Answer:
In I.9, an lvalue was defined as an expression to which a value can be assigned. It was also explained that an lvalue appears on the left side of an assignment statement. Therefore, an rvalue can be defined as an expression that can be assigned to an lvalue. The rvalue appears on the right side of an assignment statement.
Unlike an lvalue, an rvalue can be a constant or an expression, as shown here:

int x, y;
x = 1; /* 1 is an rvalue; x is an lvalue */
y = (x + 1); /* (x + 1) is an rvalue; y is an lvalue */

As stated in I.9, an assignment statement must have both an lvalue and an rvalue. Therefore, the following statement would not compile because it is missing an rvalue:

int x;
x = void_function_call() /* the function void_function_call() returns nothing */
If the function had returned an integer, it would be considered an rvalue because it evaluates into something
that the lvalue, x, can store.
Reference:
I.9: What is an lvalue?
I.10: Can an array be an lvalue?


I.12: Is left-to-right or right-to-left order guaranteed for operator precedence?
Answer:
The simple answer to this question is neither. The C language does not always evaluate left-to-right or right to- left. Generally, function calls are evaluated first, followed by complex expressions and then simple expressions. Additionally, most of today’s popular C compilers often rearrange the order in which the expression is evaluated in order to get better optimized code. You therefore should always implicitly define
your operator precedence by using parentheses.
For example, consider the following expression:

a = b + c/d / function_call() * 5

The way this expression is to be evaluated is totally ambiguous, and you probably will not get the results you want. Instead, try writing it by using implicit operator precedence:
a = b + (((c/d) / function_call()) * 5)

Using this method, you can be assured that your expression will be evaluated properly and that the compiler will not rearrange operators for optimization purposes.

I.10: Can an array be an lvalue?

Answer:
In I.9, an lvalue was defined as an expression to which a value can be assigned. Is an array an expression to which we can assign a value? The answer to this question is no, because an array is composed of several separate array elements that cannot be treated as a whole for assignment purposes. The following statement is therefore illegal:

int x[5], y[5];
x = y;
You could, however, use a for loop to iterate through each element of the array and assign values individually, such as in this example:

int i;
int x[5];
int y[5];
...
for (i=0; i<5; your_name =" my_name;">Reference:
I.9: What is an lvalue?
I.11: What is an rvalue?

I.9: What is an lvalue?

Answer:
An lvalue is an expression to which a value can be assigned. The lvalue expression is located on the left side of an assignment statement, whereas an rvalue (see FAQ I.11) is located on the right side of an assignment statement. Each assignment statement must have an lvalue and an rvalue. The lvalue expression must reference a storable variable in memory. It cannot be a constant. For instance, the following lines show a few examples of lvalues:

int x;
int* p_int;
x = 1;
*p_int = 5;

The variable x is an integer, which is a storable location in memory.

Therefore, the statement x = 1 qualifies x to be an lvalue. Notice the second assignment statement, *p_int = 5. By using the * modifier to reference the area of memory that p_int points to, *p_int is qualified as an lvalue. In contrast, here are a few examples of what would not be considered lvalues:

#define CONST_VAL 10
int x;
/* example 1 */
1 = x;
/* example 2 */
CONST_VAL = 5;

In both statements, the left side of the statement evaluates to a constant value that cannot be changed because constants do not represent storable locations in memory. Therefore, these two assignment statements do notcontain lvalues and will be flagged by your compiler as errors

I.10: Can an array be an lvalue?
I.11: What is an rvalue?

I.8: What is the difference between goto and longjmp()

Answer:
A goto statement implements a local jump of program execution, and the longjmp() and setjmp() functions implement a nonlocal, or far, jump of program execution. Generally, a jump in execution of any kind should be avoided because it is not considered good programming practice to use such statements as goto and longjmp in your program.
A goto statement simply bypasses code in your program and jumps to a predefined position. To use the goto statement, you give it a labeled position to jump to. This predefined position must be within the same function. You cannot implement gotos between functions.
Here is an example of a goto statement:

void bad_programmers_function(void)
{
int x;
printf(“Excuse me while I count to 5000...\n”);
x = 1;
while (1)
{
printf(“%d\n”, x);
if (x == 5000)
goto all_done;
else
x = x + 1;
}
all_done:
printf(“Whew! That wasn’t so bad, was it?\n”);
}

This example could have been written much better, avoiding the use of a goto statement. Here is an example of an improved implementation:
void better_function(void)
{
int x;
printf(“Excuse me while I count to 5000...\n”);
for (x=1; x<=5000; x++)
printf(“%d\n”, x);
printf(“Whew! That wasn’t so bad, was it?\n”);
}
As previously mentioned, the longjmp() and setjmp() functions implement a nonlocal goto.
When your program calls setjmp(), the current state of your program is saved in a structure of type jmp_buf. Later, your program can call the longjmp() function to restore the program’s state as it was when you called setjmp().
Unlike the goto statement, the longjmp() and setjmp() functions do not need to be implemented in the same function. However, there is a major drawback to using these functions: your program, when restored to its previously saved state, will lose its references to any dynamically allocated memory between the longjmp() and the setjmp(). This means you will waste memory for every malloc() or calloc() you have implemented between your longjmp() and setjmp(), and your program will be horribly inefficient. It is highly recommended that you avoid using functions such as longjmp() and setjmp() because they, like the goto statement, are quite often an indication of poor programming practice.
Here is an example of the longjmp() and setjmp() functions:

#include
#include
jmp_buf saved_state;
void main(void);
void call_longjmp(void);
void main(void)
{
int ret_code;
printf(“The current state of the program is being saved...\n”);
ret_code = setjmp(saved_state);
if (ret_code == 1)
{
printf(“The longjmp function has been called.\n”);
printf(“The program’s previous state has been restored.\n”);
exit(0);
}
printf(“I am about to call longjmp and\n”);
printf(“return to the previous program state...\n”);
call_longjmp();
}
void call_longjmp(void)
{
longjmp(saved_state, 1);
}

I.7: How can you tell whether a loop ended prematurely?

Answer:
Generally, loops are dependent on one or more variables. Your program can check those variables outside the loop to ensure that the loop executed properly. For instance, consider the following example:

#define REQUESTED_BLOCKS 512
int x;
char* cp[REQUESTED_BLOCKS];
/* Attempt (in vain, I must add...) to
allocate 512 10KB blocks in memory. */
for (x=0; x< REQUESTED_BLOCKS; x++)
{
cp[x] = (char*) malloc(10000, 1);
if (cp[x] == (char*) NULL)
break;
}
/* If x is less than REQUESTED_BLOCKS, the loop has ended prematurely. */
if (x < REQUESTED_BLOCKS)
printf(“Bummer! My loop ended prematurely!\n”);

Notice that for the loop to execute successfully, it would have had to iterate through 512 times. Immediately following the loop, this condition is tested to see whether the loop ended prematurely. If the variable x isanything less than 512, some error has occurred