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what will be the output:
main(){char ch;int a=10;printf("%d",ch);}
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Answer Posted / lnk
In the above coding ...
there is a char "ch " assigned with out a value
and a int" a " assigned with a value ...
and it compiles to print a interger (%d) ...
but tries to read a int value fronm ch (which does not
match the data type ) so it gives garbage value .with
compile time warning !
| Is This Answer Correct ? | 1 Yes | 0 No |
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