Answer / mickey

-128

Answer / susie

Answer :

Behavior is implementation dependent.

Explanation:

The detail if the char is signed/unsigned by default
is implementation dependent. If the implementation treats
the char to be signed by default the program will print –128
and terminate. On the other hand if it considers char to be
unsigned by default, it goes to infinite loop.

Rule:

You can write programs that have implementation
dependent behavior. But dont write programs that depend on
such behavior.

void main() { static int i=i++, j=j++, k=k++; printf(“i = %d j = %d k = %d”, i, j, k); }

3 Answers  

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How we print the table of 3 using for loop in c programing?

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Is the following statement a declaration/definition. Find what does it mean? int (*x)[10];

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Cau u say the output....?

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main() { char p[ ]="%d\n"; p[1] = 'c'; printf(p,65); }

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main() { int i = 257; int *iPtr = &i; printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) ); }

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main() { int i=5,j=10; i=i&=j&&10; printf("%d %d",i,j); }

1 Answers  

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void main() { char ch; for(ch=0;ch<=127;ch++) printf(“%c %d \n“, ch, ch); }

1 Answers  

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