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Hello Experts,

I'm trying to help my daughter with her C homeowork. I'm a longtime developer but have not worked in C in decades. She has the following problem:

The mathematical operation min(x,y) can be represented by the conditional

expression

(x < y) ? x : y

In similar fashion, using only conditional expressions, describe the

mathematical operations:

min(x, y, z) and max(x,y,z,w)

I understand how the min(x,y) translate into (x < y) ? x : y

and that (x < y) ? x : y is tha same as If x < y then x else y

but am drawing a blank as to how to code the two problems.

Can anyone help?

Thanks,

Steve

I'm trying to help my daughter with her C homeowork. I'm a longtime developer but have not worked in C in decades. She has the following problem:

The mathematical operation min(x,y) can be represented by the conditional

expression

(x < y) ? x : y

In similar fashion, using only conditional expressions, describe the

mathematical operations:

min(x, y, z) and max(x,y,z,w)

I understand how the min(x,y) translate into (x < y) ? x : y

and that (x < y) ? x : y is tha same as If x < y then x else y

but am drawing a blank as to how to code the two problems.

Can anyone help?

Thanks,

Steve

1 Solution

as you wrote:

min2(x,y) (x < y) ? x : y;

so

min3(x,y,z) min2( min2( x, y ), z );

and

max4( x, y, z, w ) max2( max2( max2( x, y ), z ), w );

hope it helps

ike

int main()

{

int i=10,j=3,k=12,l=9;

int max,min;

min = (i<j)?((i<k)?i:k):((j<k)?j

max = (j>i)?((j>k)?(j>l?j:l):(k>

printf("min %d max %d",min,max);

}

Cheers!

sunnycoder

first u hav to pass the numbers into the array

pass the array and the no. arg over the function named min

the function is like this,

int min(int arr[20],int noarg) //arr[20] is the array which are having no.//noarg is the no. arguments

{

int i,min=arr[0];

for(i=1;i<noarg;i++)

{

min=(min<arr[i])?min:arr[i

}

return min;

}

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(x<y) ? ((x < z) ? x : z) : ((y < z) ? y : z) ;

i hope this works....

the same way you can write to max(x,y,z,w)