⟩ The DBA is A. a person B. a computer device C. a communication technique D. All of the above E. None of the above
D. All of the above
D. All of the above
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]);
Can you please explain the difference between string and an array?
What will be output if you will execute following c code? #include<stdio.h> #define var 3 void main(){ char *cricket[var+~0]={"clarke","kallis"}; char *ptr=cricket[1+~0]; printf("%c",*++ptr);
What will be output if you will execute following c code? #include<stdio.h> void main(){ long myarr[2][4]={0l,1l,2l,3l,4l,5l,6l,7l}; printf("%ldt",myarr[1][2]); printf("%ld%ldt",*(myarr[1]+3),3[myarr[1]]); printf("%ld%ld%ldt" ,*(*(myarr+1)+2),*(1[myarr]+2),3[1[myarr]]);
What will be output if you will execute following c code? #include<stdio.h> enum power{ Dalai, Vladimir=3, Barack, Hillary }; void main(){ float leader[Dalai+Hillary]={1.f,2.f,3.f,4.f,5.f}; enum power p=Barack; printf("%0.f",leader[p>>1+1]);