Which of the following functions find the count of even numbers enclosed within double quotes in an array with 7 ? a. int countEven( char *a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump+=1%;}}return nump;} b. int countEven( char *a[]) { int i.j,nump = 0;for (j = 0; j < 7; ++j) { if(atoi(a[j])%2=%=D0){nump++3;} }return nump;} c. int countEven( char *a[]) { int ij,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j-1])%2=%=D0){nump++;} }return nump;} %3D d. int countEven( char *a[]) { int i,j.nump 0; for (j = 0; j< 7; j++) { if(atoi(a[j])%2==0){nump+%3Dnump;} }return nump;} e. int countEven( char *a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump++;} }return nump;}

Which of the following functions find the count of even numbers enclosed within double quotes in an array with 7 ? a. int countEven( char a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump+=1%;}}return nump;} b. int countEven( char a[]) { int i.j,nump = 0;for (j = 0; j < 7; ++j) { if(atoi(a[j])%2=%=D0){nump++3;} }return nump;} c. int countEven( char a[]) { int ij,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j-1])%2=%=D0){nump++;} }return nump;} %3D d. int countEven( char a[]) { int i,j.nump 0; for (j = 0; j< 7; j++) { if(atoi(a[j])%2==0){nump+%3Dnump;} }return nump;} e. int countEven( char *a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump++;} }return nump;}

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

Which of the following functions find the count of even numbers enclosed within double quotes in an array with 7 ?
a. int countEven( char *a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump+=1%;}}return
nump;}
b. int countEven( char *a[]) { int i.j,nump = 0;for (j = 0; j < 7; ++j) { if(atoi(a[j])%2=%=D0){nump++3;} }return
nump;}
c. int countEven( char *a[]) { int ij,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j-1])%2=%=D0){nump++;} }return
nump;}
%3D
d. int countEven( char *a[]) { int i,j.nump 0; for (j = 0; j< 7; j++) { if(atoi(a[j])%2==0){nump+%3Dnump;}
}return nump;}
e. int countEven( char *a[]) { int i,j,nump = 0;for (j = 0; j < 7; j++) { if(atoi(a[j])%2!=0){nump++;} }return
nump;}

Expert Solution

Step 1

The correct option, that will count the even numbers enclosed within double quotes in an array, is b.

int countEven(char *a[])

{

int i, j, nump = 0;

for (j = 0; j < 7; j++)

{

if (atoi(a[j]) % 2 == 0)

{

nump++;

}

return nump;

}

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