What causes the reappointment of this integer pointer?

Question

I am new to C and I have this question. why the following code failure occurs:

int *a = 10;
*a = 100;

Answer 1

Because you are trying to write 100 to memory location 0x0000000A, which is probably not allocated to your program. I.e

int *a = 10;

does not mean that the pointer "a" points to a place in memory with a value of 10. This means that it points to address 10 (0x0000000A) in memory. Then you want to write something to this address, but you do not have “rights” to do this, since it does not stand out

You can try the following:

int *a = malloc(sizeof(int));
*a = 100;

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Answer 2

, (10).

int cell = 10;
int *a = &cell; // a points to address of cell
*a = 100;       // content of cell changed

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Answer 3

malloc(), , int. :

a = malloc(sizeof(int));

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a = malloc(sizeof *a);

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Answer 4

- a. .

int *a = NULL;

a = malloc (sizeof (int));

if (a != NULL)
{
*a =10;
}

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int a* = NULL;
int b = 10;

a = &b;

, -

*a = 100;

, b == 100

: http://home.netcom.com/~tjensen/ptr/pointers.pdf

Answer 5

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int *a = 10;

a. a 10.

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*a = 100;

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Answer 6

int, 10 (), int . 10 , . :

int *a;
a = malloc(sizeof(int));
*a = 10;
printf("a=%i\n", *a);
free(a);

Answer 7

? 10 int *, :

int *a = (int *) 10;
*a = 100;

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Answer 8

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Answer 9

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(int*) a = 10;
(*a) = 100;

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(void*) v = NULL;

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struct Hello {
    int id;
    char* name;
};

...

struct Hello* hello_ptr = malloc(sizeof Hello);
hello_ptr->id = 5;
hello_ptr->name = "Cheery";

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void* malloc(size_t size);

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free(hello_ptr);

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Answer 10

:

int* a = 10;
*a = 100;

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:

"Pointer-to-int 'a' becomes 10"
"Value-pointed-to-by 'a' becomes 100"

:

"Value-pointed-to-by 10 becomes 100"

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:

int* ptr = (int*)10;  // You've guessed at a memory address, and probably got it wrong
int* ptr = malloc(sizeof(int)); // OS gives you a memory address at runtime  

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