1. What will be the output of the following programs. a) #include <stdio.h> Main() { Int x=4; While(x==1) { X=x-1; Printf(“%d”,x); --x; } }

7 Answers   CSC,

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about c language

0 Answers  

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Explain how do I determine whether a character is numeric, alphabetic, and so on?

0 Answers  

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where do we use volatile keyword?

1 Answers  

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What is page thrashing?

0 Answers  

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int main(){ float f=8.0; if(f==8.0) printf("good"); else printf("bad"); } what is the answere and explain it?

3 Answers  

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Are enumerations really portable?

0 Answers  

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in b=6.6/a+(2*a+(3*c)/a*d)/(2/n); which operation will be performed first a) 6.6/a b) 2*a c) 3*c d) 2/n

1 Answers  

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main() { intx=2,y=6,z=6; x=y=z; printf(%d",x) }

5 Answers   Amazon, HCL, Thought Works,

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void main() { int s[4][2]={ {1234,56},{1212,33},{1434,80},{1312,78} }; int (*p)[2]; int i,j,*pint; for(i=0;i<=3;i++) { p=&s[i]; pint=p; printf("\n"); for(j=0;j<=1;j++) printf("%d",*(pint+j)); } } while running this program it shows a warning-suspicious pointer conversion ie pint=p; my que is why should we assign the value of p to pint again.why cant we use it directly as *(p+j)..but if i use like tat the o/p is garbage value..

1 Answers  

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1234554321 1234 4321 123 321 12 21 1 1 12 21 123 321 1234 4321 1234554321

0 Answers  

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class foo { public: static int func(const char*& p) const; }; This is illegal, why?

8 Answers   Google,

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