Questions for: Functions
#include<stdio.h>
int check(int);
int main()
{
int i=45, c;
c = check(i);
printf("%d\n", c);
return 0;
}
int check(int ch)
{
if(ch >= 45)
return 100;
else
return 10;
}
Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value.
Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45.
The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10.
Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100)
Step 4: printf("%d\n", c); It prints the value of variable c.
Hence the output of the program is '100'.
#include<stdio.h>
int addmult(int ii, int jj)
{
int kk, ll;
kk = ii + jj;
ll = ii * jj;
return (kk, ll);
}
int main()
{
int i=3, j=4, k, l;
k = addmult(i, j);
l = addmult(i, j);
printf("%d, %d\n", k, l);
return 0;
}Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.
The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d\n", k, l); It prints the value of k and l
Hence the output is "12, 12".
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#include<stdio.h>
int func1(int);
int main()
{
int k=35;
k = func1(k=func1(k=func1(k)));
printf("k=%d\n", k);
return 0;
}
int func1(int k)
{
k++;
return k;
}
Step 1: int k=35; The variable k is declared as an integer type and initialized to 35.
Step 2: k = func1(k=func1(k=func1(k))); The func1(k) increement the value of k by 1 and return it. Here the func1(k) is called 3 times. Hence it increements value of k = 35 to 38. The result is stored in the variable k = 38.
Step 3: printf("k=%d\n", k); It prints the value of variable k "38".
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#include<stdio.h>
int i;
int fun1(int);
int fun2(int);
int main()
{
extern int j;
int i=3;
fun1(i);
printf("%d,", i);
fun2(i);
printf("%d", i);
return 0;
}
int fun1(int j)
{
printf("%d,", ++j);
return 0;
}
int fun2(int i)
{
printf("%d,", ++i);
return 0;
}
int j=1;
Step 1: int i; The variable i is declared as an global and integer type.
Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.
Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.
Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.
Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.
Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.
Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.
Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Hence the output is "4 3 4 3".
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#include<stdio.h>
int addmult(int ii, int jj)
{
int kk, ll;
kk = ii + jj;
ll = ii * jj;
return (kk, ll);
}
int main()
{
int i=3, j=4, k, l;
k = addmult(i, j);
l = addmult(i, j);
printf("%d %d\n", k, l);
return 0;
}Discuss About this Question.
Discuss About this Question.