Look at the following code. int x = 7; int *iptr = & x; what will be displayed if you send the expression *iptr to cout ? what happens if you send theexpression ptr to cout ?

The answer to the given question can be given as:

The value of *iptr is 7. and the value of iptr is dynamic.

Explanation:

In the c++ code, it is defined that x is an integer variable that assigns a value which is 7. Then we define a pointer variable that is *iptr. This variable holds an address of the x variable. When we print the value of the iptr variable. if we use the expression *iptr to print value of the pointer variable by cout that is used in c++ for pint values. so the value of the iptr is 7. If we use the expression iptr sent to cout so we show the address of the variable x. In the pointer, it manages the addresses of dynamically allocated so the address of the variable is changed on execution time.

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