Analyze and determine the outputs of the following program statements: #include #include int main() ( float a-3, c4,"p, "q; int i,j,k,n-3, b-3,w=2; float x[3] [4]-(9,8,1,2,3,4,5,6,7,8,9,10), z[4][3]; float y[4]-(1,2,3,4); for(i=0;i<4;i++) { for(j=0; j<3; j++) } z[2][1]*[2][1]; q=&z[2][1]; printf("q- %.2f\n", *q*2); z[1][1]-x[1][1]; for(i=0;i<3;++1) {c+=y[1] -1; if(w<10) (p-&n; ) else ( p+=; } printf("a-%0.2f\n c-%0.2f\n z[2][1]-x0.2f \n", a,c, z[2][1]); 1-4;j-3;k-n-1; if (il-jj-3) (1=k;=j+1;} else if(i<8) (wen+1+1+1+y[1]; wen+a+k+2+y[1];) printf("%d",w+5); } a+my[i]+b; printf("w-%d\n",w-5);)

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
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Analyze and determine the outputs of the following program statements:
#include <stdio.h>
#include <math.h>
int main()
(
float a-3, c-4,"p, *q;
int i, j, k,n-3, b=3,w=2;
float x[3] [4]=(9,8,1,2,3,4,5,6,7,8,9,10), z[4][3];
float y[4]-(1,2,3,4);
for(i=0;i<4;i++)
{ for(j=0; j<3;j++)
}
z[2][1]=x[2][¹];
q=&z[2][1];
printf("*q= %.2f\n", *q*2);
z[j][i]=x[i][j];
for(i=0;i<3; ++i)
{c+=y[1] -1;
if(w<18)
(p-&n;
}
else
{
p+=1;
printf("a-%0.2f\n c-%8.2f\n z[2][1] -%0.2f \n",a,c,z[2][1]);
i=4;j-3;k-n-1;
if (il-jj-3)
(1=k;=j+1;}
else if(i<8)
(w=n+1+1+1+y[1]; w=n+a+k+2+y[1];}
printf("w-%d",w+5);
}
a+my[i]+b;
}
printf("w-%d\n",w-5);}
Transcribed Image Text:Analyze and determine the outputs of the following program statements: #include <stdio.h> #include <math.h> int main() ( float a-3, c-4,"p, *q; int i, j, k,n-3, b=3,w=2; float x[3] [4]=(9,8,1,2,3,4,5,6,7,8,9,10), z[4][3]; float y[4]-(1,2,3,4); for(i=0;i<4;i++) { for(j=0; j<3;j++) } z[2][1]=x[2][¹]; q=&z[2][1]; printf("*q= %.2f\n", *q*2); z[j][i]=x[i][j]; for(i=0;i<3; ++i) {c+=y[1] -1; if(w<18) (p-&n; } else { p+=1; printf("a-%0.2f\n c-%8.2f\n z[2][1] -%0.2f \n",a,c,z[2][1]); i=4;j-3;k-n-1; if (il-jj-3) (1=k;=j+1;} else if(i<8) (w=n+1+1+1+y[1]; w=n+a+k+2+y[1];} printf("w-%d",w+5); } a+my[i]+b; } printf("w-%d\n",w-5);}
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