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Casting result from malloc to char (not char *) - why doesn't the compiler complain?

tmpString = (char*)malloc((strlen(name) + 1) * sizeof(char));
tmpString = (char )malloc((strlen(name) + 1) * sizeof(char));

What is the difference between these two lines?

I understand that the second line is incorrect, but for some reason the compiler is not saying anything.

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4 answers

The first line indicates the pointer (void), which malloc returns to a pointer to char, thereby preserving it as its pointer. All this tells the compiler that "memory at location X should be treated as an array of characters."

The second throw turns the pointer returned by malloc into one character. This is bad for several reasons:

  • You are losing the pointer because you just turned the pointer into something completely different.
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