int a = 5;int b = 6;int nain(void){printf("%d Xd \n", a, b);int a;printf("d Xd \n", a, b);return e;When this program runs it prints (at least using gcc -std=c99 program.c -o program on Vera Virtual).5 6;e 6;This shows that the compiler implements the second int a; by initializing the local variable a to zero. Now consider this program:int a = 5;int b = 6;int nain(void){int a-a + b;printf("d Xd \n", a, b);return e;When this second program runs it prints, perhaps surprisingly:6 6;Based on the first program, which a (local or global) is being used to compute a + b ? Finally, consider this third program:int a = 6;int b = 5;int nain(void){b = a + a;a = b + b;int a = a + b;int b= a - b;printf("d\n", a + b);return e;Activate WindowsWhat integer does this third program print to the console? (at least using gcc -std=c99 program.c -o

Given: The output of the first program was 5 6 0 6 this is because when first time the value of a…

int a 5; int b = 6; int nain(void){ printf("d Xd \n", a, b); int a; printf("d Xd \n", a, b); return e; When this program runs it prints (at least using gcc -std=c99 program.c -o program on Vera Virtual). 5 6; e 6; This shows that the compiler implements the second int a; by initializing the local variable a to zero. Now consider this program: int a = 5; int b = 6; int nain(void){ int aa+ b; printf("d d \n", a, b); return e; When this second program runs it prints, perhaps surprisingly: 6 6; Based on the first program, which a (local or global) is being used to compute a + b ? Finally, consider this third program: int a 6; int b = 5; int main(void){ b = a+ a; a = b+ b; int aa+ b; int b= a- b; printf("d\n", a + b); return e; Activate Windows What integer does this third program print to the console? (at least using gcc -std=c99 program.c -o

int a 5; int b = 6; int nain(void){ printf("d Xd \n", a, b); int a; printf("d Xd \n", a, b); return e; When this program runs it prints (at least using gcc -std=c99 program.c -o program on Vera Virtual). 5 6; e 6; This shows that the compiler implements the second int a; by initializing the local variable a to zero. Now consider this program: int a = 5; int b = 6; int nain(void){ int aa+ b; printf("d d \n", a, b); return e; When this second program runs it prints, perhaps surprisingly: 6 6; Based on the first program, which a (local or global) is being used to compute a + b ? Finally, consider this third program: int a 6; int b = 5; int main(void){ b = a+ a; a = b+ b; int aa+ b; int b= a- b; printf("d\n", a + b); return e; Activate Windows What integer does this third program print to the console? (at least using gcc -std=c99 program.c -o

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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Question

int a = 5;
int b = 6;
int nain(void){
printf("%d Xd \n", a, b);
int a;
printf("d Xd \n", a, b);
return e;
When this program runs it prints (at least using gcc -std=c99 program.c -o program on Vera Virtual).
5 6;
e 6;
This shows that the compiler implements the second int a; by initializing the local variable a to zero. Now consider this program:
int a = 5;
int b = 6;
int nain(void){
int a-a + b;
printf("d Xd \n", a, b);
return e;
When this second program runs it prints, perhaps surprisingly:
6 6;
Based on the first program, which a (local or global) is being used to compute a + b ? Finally, consider this third program:
int a = 6;
int b = 5;
int nain(void){
b = a + a;
a = b + b;
int a = a + b;
int b= a - b;
printf("d\n", a + b);
return e;
Activate Windows
What integer does this third program print to the console? (at least using gcc -std=c99 program.c -o

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