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int a[10]; char * s = "hello"; int f(int i, int b[]) { int j=i; A:{ int i=j; char c = b[i]; // ...; } return 0; } void g(char * s) { char c = s [ 0]; B:{ int a[5];: // ...; } } main () { int x=1; x = f(x,a); g(s); return 0; }

int a[10]; char * s = "hello"; int f(int i, int b[]) { int j=i; A:{ int i=j; char c = b[i]; // ...; } return 0; } void g(char * s) { char c = s [ 0]; B:{ int a[5];: // ...; } } main () { int x=1; x = f(x,a); g(s); return 0; }

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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(Run-time environment) Draw a possible organization for the runtime environment of the following C program, similar to the given figure below. 

1. after entry into block A in function f.
2. after entry into block B in function g.

PLEASE SHOW IT CLEARLY WITH THE EXPLANATION.

THANK YOU.

X: 15 Y: 10 Globalstatic area Activation record of main u: 15 V: 10 control link return address Activation record of first call to gcd u: 10 V: 5 - control link Activation record of second call to gcd return address u: 5 V: 0 Activation record of control link return address third call to gcd Direction of stack growth
Transcribed Image Text:X: 15 Y: 10 Globalstatic area Activation record of main u: 15 V: 10 control link return address Activation record of first call to gcd u: 10 V: 5 - control link Activation record of second call to gcd return address u: 5 V: 0 Activation record of control link return address third call to gcd Direction of stack growth
int a[10]; 2 || char * s = " hello"; 1 3 |int f(int i, int b[]) 5|{ int j=i; A:{ int i=j ; char c = b[i]; // ...; } return 0; 4 6 8 10 } 11 12 13 void g(char * s) { char c = s [0]; B:{ int a[5]; // ...; } 14 15 16 17 18 } 19 main () { int x=1; x = f(x, a); g(s); return 0; 20 21 22 23 24 } 25
Transcribed Image Text:int a[10]; 2 || char * s = " hello"; 1 3 |int f(int i, int b[]) 5|{ int j=i; A:{ int i=j ; char c = b[i]; // ...; } return 0; 4 6 8 10 } 11 12 13 void g(char * s) { char c = s [0]; B:{ int a[5]; // ...; } 14 15 16 17 18 } 19 main () { int x=1; x = f(x, a); g(s); return 0; 20 21 22 23 24 } 25
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