using the formula:

i=2 + (y+0.5x)

Write a program which will produce a table of values of i, y and x, where y varies from 1 to 6,

and, for each value of y, x varies from 5.5 to 12.5 in steps of 0.5 .

**#include<stdio.h>**

main()

**{**

**int**y;

**float**i,x;

**for**(y=1; y<=6; y++)

**{**

**for**(x=5.5; x<=12.5;x+=0.5)

**{**

i=2 + (y+ (0.5*x)) ;

printf("\n i=%f, y=%d, x=%f", i, y, x);

**continue**;

**}**

**}**

**}**

The file can be found at:

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## 3 comments:

Why the use of continue?

try this code its not wrking

#include

#include

int main()

{

float i,x,y;

//x=5.5;

int j;

for(j=1;j<=6;j++)

{

for(x=5.5;x<=12.5;x+=5.5)

{

i=2+(j+0.5x);

//x=x+0.5;

printf("%f\n",i);

continue;

}

}

getch();

return 0;

}

main()

{

float i,x;

int j;

for(j=1;j<=6;j++)

{

for(x=5.5;x<=12.5;x+=0.5) //x=x+0.5;

{

i=2+(j+0.5*x);

printf("%f %d %f \n",i, j, x);

continue;

}

}

getch();

return 0;

}

Now code will work

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