Malloc vs array in C

Question

I am taking a course from the MOOC CS50 from Harvard. The last lecture I read was about allocating memory and pointers (two completely new concepts).

It has been shown that it malloc(10*sizeof(char))allocates enough bytes on the heap to store 10 characters and returns a pointer to the first byte, which can be stored in another variable as follows char *x = malloc(10*sizeof(char)). To free memory you must use free(x).

But there is another way to make the computer reserve enough memory to store 10 characters, i.e. char c[10].

• Is there a ctype pointer in the code snippet above char*?
• Does heapchar c[10] memory save how malloc?
• Ways to allocate equivalent memory?
• char c[3] = "aaa";free(c);returns a runtime error; therefore it seems that I cannot free the memory that I allocated with char c[3]. Why is this?

I would really appreciate an answer like the one who just found out about pointers.

Answer 1

What was taught is that it malloc(10*sizeof(char))allocates enough bytes on the heap to store 10 characters and returns a pointer to the first byte, which can be stored in another variable as follows char *x = malloc(10*sizeof(char)). To free memory, use free(x).

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printf("sizeof(x) = %zd\n", sizeof(x));
printf("sizeof(*x) = %zd\n", sizeof(*x));
printf("sizeof(c) = %zd\n", sizeof(c));

sizeof(x) = 8
sizeof(*x) = 1
sizeof(c) = 10

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