a) delayed until run time
b) preponed to compile time
c) preponed to load time
d) none of the mentioned
Answer: a
Explanation: If the process can be moved during its execution from one memory segment to another, then binding must be delayed until run time
Related Posts
What will be the output of the following C code?
#include < stdio.h >
int main()
{
int i = 0;
for (foo(); i == 1; i = 2)
printf(“In for loop\n”);
printf(“After loop\n”);
}
int foo()
{
return 1;
}What will be the output of the following C code?
#include < stdio.h >
int main()
{
int i = 0;
for (i++; i == 1; i = 2)
printf(“In for loop “);
printf(“After loop\n”);
}What will be the output of the following C code?
#include < stdio.h >
int main()
{
int i = 0;
for (; ; ;)
printf(“In for loop\n”);
printf(“After loop\n”);
}What will be the output of the following C code?
#include < stdio.h >
void main()
{
int k;
for (k = -3; k < -5; k++)
printf(“Hello”);
}What will be the output of the following C code?
#include < stdio.h >
void main()
{
double k = 0;
for (k = 0.0; k < 3.0; k++);
printf(“%lf”, k);
}What will be the output of the following C code?
#include < stdio.h >
void main()
{
double k = 0;
for (k = 0.0; k < 3.0; k++)
printf(“Hello”);
}What will be the output of the following C code?
#include < stdio.h >
void main()
{
int k = 0;
for (k < 3; k++)
printf(“Hello”);
}
Join The Discussion