main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer / susie
Answer :
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or
0 . And also the logical AND (&&) operator has higher
priority over the logical OR (||) operator. So the
expression ‘i++ && j++ && k++’ is executed first. The
result of this expression is 0 (-1 && -1 && 0 = 0). Now
the expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for ‘0 || 0’ combination- for
which it gives 0). So the value of m is 1. The values of
other variables are also incremented by 1.
Is This Answer Correct ? | 39 Yes | 11 No |
To Write a C program to remove the repeated characters in the entered expression or in entered characters(i.e) removing duplicates.
19 Answers Amazon, BITS, Microsoft, Syncfusion, Synergy, Vector,
Declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
# include <stdio.h> int one_d[]={1,2,3}; main() { int *ptr; ptr=one_d; ptr+=3; printf("%d",*ptr); }
What is the hidden bug with the following statement? assert(val++ != 0);
main() { float me = 1.1; double you = 1.1; if(me==you) printf("I love U"); else printf("I hate U"); }
void main() { if(~0 == (unsigned int)-1) printf(“You can answer this if you know how values are represented in memory”); }
main() { static int a[3][3]={1,2,3,4,5,6,7,8,9}; int i,j; static *p[]={a,a+1,a+2}; for(i=0;i<3;i++) { for(j=0;j<3;j++) printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j), *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i)); } }
int main() { int x=10; printf("x=%d, count of earlier print=%d", x,printf("x=%d, y=%d",x,--x)); getch(); } ================================================== returns error>> ld returned 1 exit status =================================================== Does it have something to do with printf() inside another printf().
what is oop?
main() { char *p; printf("%d %d ",sizeof(*p),sizeof(p)); }
void main() { int i; char a[]="\0"; if(printf("%s\n",a)) printf("Ok here \n"); else printf("Forget it\n"); }
main( ) { int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}}; printf(“%u %u %u %d \n”,a,*a,**a,***a); printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1); }