⟩ _____ means that the manager spends his or her time dealing with exceptions or those situations which are out of control A. relevant B. management by exception C. predictive reports D. control E. None of the above
B. management by exception
B. management by exception
What will be output if you will execute following c code? #include<stdio.h> union group{ char xarr[2][2]; char yarr[4]; }; void main(){ union group x={'A','B','C','D'}; printf("%c",x.xarr[x.yarr[2]-67][x.yarr[3]-67]); }
Can you please explain the difference between array_name and &array_name?
What will be output if you will execute following c code? #include<stdio.h> #include<math.h> typedef struct{ char *name; double salary; }job; void main(){ static job a={"TCS",15000.0}; static job b={"IBM",25000.0}; static job c={"Google",35000.0}; int x=5; job * arr[3]={&a,&b,&c}; printf("%s %ft",(3,x>>5-4)[*arr]); } double myfun(double d){ d-=1; return d;
What will be output if you will execute following c code? #include<stdio.h> #include<math.h> double myfun(double); void main(){ double(*array[3])(double); array[0]=exp; array[1]=sqrt; array[2]=myfun; printf("%.1ft",(*array)((*array[2])((**(array+1))(4)))); } double myfun(double d){ d-=1; return d; }
What will be output if you will execute following c code? #include<stdio.h> void main(){ int array[2][3]={5,10,15,20,25,30}; int (*ptr)[2][3]=&array; printf("%dt",***ptr); printf("%dt",***(ptr+1)); printf("%dt",**(*ptr+1)); printf("%dt",*(*(*ptr+1)+2));
Tell me why can't constant values be used to define an array's initial size
What will be output if you will execute following c code? #include<stdio.h> void main(){ static int a=2,b=4,c=8; static int *arr1[2]={&a,&b}; static int *arr2[2]={&b,&c}; int* (*arr[2])[2]={&arr1,&arr2}; printf("%d %dt",*(*arr[0])[1], *(*(**(arr+1)+1)));
What will be output if you will execute following c code? #include<stdio.h> #define var 3 void main(){ char *ptr="cquestionbank"; printf("%d",-3[ptr]);
What will be output if you will execute following c code? #include<stdio.h> #define var 3 void main(){ short num[3][2]={3,6,9,12,15,18}; printf("%d %d",*(num+1)[1],**(num+2));
What will be output if you will execute following c code? #include<stdio.h> #define var 3 void main(){ char data[2][3][2]={0,1,2,3,4,5,6,7,8,9,10,11}; printf("%o",data[0][2][1]);