Dennis Lee Bieber <[EMAIL PROTECTED]> wrote: > Of course, there is also the confusion between "type cast" and "type > conversion" -- at least, for me...
> cast taking the bit-pattern of a value in one "type" and > interpreting the same bit-pattern as if it were a different "type" > conversion taking the value of a bit-pattern in one "type" and > converting it to the bit pattern of the equivalent value in another > "type" >From where have you learned those definitions? If it's from C, then you have read the wrong books. A cast in C is a type conversion, with the same semantics as you write under "conversion" above. The C standard (ISO/IEC 9899:1999) says: 6.5.4 Cast operators [...] Semantics Preceding an expression by a parenthesized type name converts the value of the expression to the named type. This construction is called a cast. A cast that specifies no conversion has no effect on the type or value of an expression. ("A cast that specifies no conversion" refers to when you cast from one type to the same type, e.g. '(int)x' if x is of the type 'int'.) You may also try this program: #include <stdio.h> #include <string.h> int main(void) { /* Assumption: sizeof(float)==sizeof(int). This is the most * common case on modern computers. */ float f = -17.0; int i = -23; float fjunk; int ijunk; printf("Cast: %d %10.6f\n", (int)f, (float)i); memcpy(&fjunk, &i, sizeof i); memcpy(&ijunk, &f, sizeof f); printf("Bitcopy: %d %10.6f\n", ijunk, fjunk); return 0; } If a cast had been what you believed, both printf() statements above would give the same output. Unless your C compiler uses some really strange floating point representation, they will print rather different things. The first one must print Cast: -17 -23.000000 showing very clearly that a cast is a type conversion. The second printf() will print some seemingly random numbers, showing that those bit-patterns will be interpreted very differently when interpreted as another type. What you might have been confused by, is dereferencing a casted pointer. Add the following statement: printf("Pointer cast: %d %10.6f\n", *(int*)&f, *(float*)&i); to the program. It should output the same numbers as the "Bitcopy" printf(). But what is cast here is the *address* of the variables, not the actual contents of them. It is the *dereferencing* of those casted pointers that interpret the bit patterns in the variables as if they were another type, not the casting itself. -- Thomas Bellman, Lysator Computer Club, Linköping University, Sweden "I don't think [that word] means what you ! bellman @ lysator.liu.se think it means." -- The Princess Bride ! Make Love -- Nicht Wahr!
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