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#define DCHAR char*
typedef char* TCHAR;
if using these following variables will be declared like
DCHAR ch1, ch2;
TCHAR ch3, ch4;
then what will be types of ch1, ch2, ch3 and ch4?
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Answer / shiva
ch1,ch3 and ch4 are char* types whiel ch2 is char type.
Please let me know if I am wrong.
| Is This Answer Correct ? | 22 Yes | 5 No |
Answer / james holdcroft
Explanation:
DCHAR is a macro, which is expanded by the preprocessor.
TCHAR is a synonym for a pointer-to-char type.
After preprocessing, the code fragment expands to:
char* ch1, ch2;
TCHAR ch3, ch4;
which is perhaps more clearly written as:
char *ch1, ch2;
TCHAR ch3, ch4;
which is equivalent to:
char *ch1; char ch2;
char *ch3; char *ch4;
| Is This Answer Correct ? | 4 Yes | 1 No |
Answer / ashish
ch1 and ch2 are the char * and ch2 and ch4 are char type
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / movemaker
^^All 4 will be char* types, why the hell would ch2 be char type like said above by Shiva . :P .
| Is This Answer Correct ? | 1 Yes | 7 No |
#include<stdio.h> int main() { int i=0,j=1,k=2,m,n=0; m=i++&&j++&&k++||n++; printf("%d,%d,%d,%d,%d",i,j,k,m,n); }
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