Predict te gut of the following snippet.
int main()
{
int x = 4.*y:
y = &x;(*y)++;
printf("%d", *y);
return 0;
}
Answers
Answer:
What will be the output of the following C code: #include <stdio.h> void m (int *p) {int I = 0; for (I = 0;I < 5; I++) printf ("%d\t", p[I]);} void main() {int a [5] = {6, 5, 3}; m(&a);}?
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What will be the output of the following C code: #include <stdio.h> void m (int *p) {int I = 0; for (I = 0;I < 5; I++) printf ("%d\t", p[I]);} void main() {int a [5] = {6, 5, 3}; m(&a);}?
Reformatted:
#include <stdio.h>
void m (int *p) {
int I = 0;
for (I = 0; I < 5; I++)
printf ("%d\t", p[I]);
}
void main() {
int a [5] = {6, 5, 3};
m(&a);
}
I see two major problems and one minor one.
void main() should be int main(void).
If your textbook, tutorial, or instructor tells you to use void main(), find a better one.
m(&a) should be m(a). Your function m expects an argument of type int*. &a is of type int(*)[5] (pointer to array of 5 int). a is an array, but it “decays” to a pointer in most contexts.
Minor: The output doesn’t end with a newline. Add something like putchar('\n') at the end of m or at the end of main.
With those problems fixed, the output is well defined. You should be able to compile and run the program for yourself and see what it does, so I won’t bother to tell you what the output is (and if you’re not able to compile and run C programs, knowing what the output is won’t do you much good). I’ll just mention that, contrary to some of the other answers, the values of a[3] and a[4] are well defined.
Answer:
(i) When the value of imports is more than the value of exports, it is called 'unfavorable balance of trade. (ii) When the value of exports is more than the value of imports, it is called 'favourable balance of trade.
Explanation:
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