Answer / susie

Answer :

infinite loop

Explanation

The difference between the previous question and this
one is that the char is declared to be unsigned. So the i++
can never yield negative value and i>=0 never becomes false
so that it can come out of the for loop.

main() { int a=2,*f1,*f2; f1=f2=&a; *f2+=*f2+=a+=2.5; printf("\n%d %d %d",a,*f1,*f2); }

6 Answers  

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What is full form of PEPSI

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main( ) { char *q; int j; for (j=0; j<3; j++) scanf(“%s” ,(q+j)); for (j=0; j<3; j++) printf(“%c” ,*(q+j)); for (j=0; j<3; j++) printf(“%s” ,(q+j)); }

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How can you relate the function with the structure? Explain with an appropriate example.

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main() { int i=0; for(;i++;printf("%d",i)) ; printf("%d",i); }

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Write a single line c expression to delete a,b,c from aabbcc

2 Answers   Microsoft,

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main() { unsigned int i=10; while(i-->=0) printf("%u ",i); }

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Write a program to implement the motion of a bouncing ball using a downward gravitational force and a ground-plane friction force. Initially the ball is to be projected in to space with a given velocity vector

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main() { printf("\nab"); printf("\bsi"); printf("\rha"); }

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There were 10 records stored in “somefile.dat” but the following program printed 11 names. What went wrong? void main() { struct student { char name[30], rollno[6]; }stud; FILE *fp = fopen(“somefile.dat”,”r”); while(!feof(fp)) { fread(&stud, sizeof(stud), 1 , fp); puts(stud.name); } }

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How to access command-line arguments?

4 Answers  

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main() { int i=5; printf(“%d”,i=++i ==6); }

1 Answers  

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