438Chapter 9 ▼ Structures and Unions3 The program in the previous exercise used a new block to illustrate a certain point.Because of this, the code was not very natural. In this exercise, we use a typedefagain, and this time the code is quite straightforward.typedef char *string;int main(void){string a []b[]={"I", "like", "to", "fight,"},{"pinch,", "and", "bite."};printf("%s %s %s %s %s %s %s\n",a[0], a[1], a[2], a[3], b[0], b[1], b[2]);return 0;}Execute the program so that you understand its effects. If the typedef is replacedby the line#define string char *then the program will not compile. Explain why the typedef works, but the#define does not. To make the program work with the #define, a single characterneeds to be added. What is the character?

Answer is given below with explanation

3 The program in the previous exercise used a new block to illustrate a certain point. Because of this, the code was not very natural. In this exercise, we use a typedef again, and this time the code is quite straightforward. typedef char * string; int main(void) { string a [] = b[] = {"I", "like", "to", "fight,"}, {"pinch," "and", "bite."}:

3 The program in the previous exercise used a new block to illustrate a certain point. Because of this, the code was not very natural. In this exercise, we use a typedef again, and this time the code is quite straightforward. typedef char * string; int main(void) { string a [] = b[] = {"I", "like", "to", "fight,"}, {"pinch," "and", "bite."}:

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept of Pointer Parameter

Pointers are the special variable that stores the variable address, instead of storing the variable value. Pointer is a powerful feature of a programming language. A pointer is mostly used to save memory space and improve execution time.

Question

438
Chapter 9 ▼ Structures and Unions
3 The program in the previous exercise used a new block to illustrate a certain point.
Because of this, the code was not very natural. In this exercise, we use a typedef
again, and this time the code is quite straightforward.
typedef char *
string;
int main(void)
{
string a []
b[]
=
{"I", "like", "to", "fight,"},
{"pinch,", "and", "bite."};
printf("%s %s %s %s %s %s %s\n",
a[0], a[1], a[2], a[3], b[0], b[1], b[2]);
return 0;
}
Execute the program so that you understand its effects. If the typedef is replaced
by the line
#define string char *
then the program will not compile. Explain why the typedef works, but the
#define does not. To make the program work with the #define, a single character
needs to be added. What is the character?

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