C printf 打印一个我没有要求的数组

Question

I have recently started learning C and I got into this problem where printf() prints an array I didn't ask for. I was expecting an error since I used %s format in char array without the '\0', but below is what I got.

char testArray1[] = { 'a','b','c'};
char testArray2[] = { 'q','w','e','r','\0' };

printf("%c", testArray1[0]);
printf("%c", testArray1[1]);
printf("%c\n", testArray1[2]);

printf("%s\n", testArray1);

the result is

abc
abcqwer

thanks

Answer 1

The format "%s" expects that the corresponding argument points to a string: sequence of characters terminated by the zero character '\0' .

printf("%s\n", testArray1);

As the array testArray1 does not contain a string then the call above has undefined behavior.

Instead you could write

printf("%.*s\n", 3,testArray1);

or

printf("%.3s\n", testArray1);

specifying exactly how many elements of the array you are going to output.

Pay attention to that in C instead of these declarations

char testArray1[] = { 'a','b','c'};
char testArray2[] = { 'q','w','e','r','\0' };

you may write

char testArray1[3] = { "abc" };
char testArray2[] = { "qwer" };

or that is the same

char testArray1[3] = "abc";
char testArray2[] = "qwer";

In C++ the first declaration will be invalid.

Answer 2

Add zero at the end of first array:

char testArray1[] = { 'a','b','c', 0 };

Otherwise printf continues with memory after 'c' until zero byte and there is the second array.

PS: zero 0 is 100% identical to longer ASCII '\0' .

Answer 3

%s indeed stop when encountered \0 , but testArray1 didn't have that \0 , so it keeps printing the following bytes in the memory.

And the compiler magically(actually intentionally) places the testArray2 next to testArray1 , the memory is like:

a b c q w e r \0
^                testArray1 starts here
      ^          testArray2 starts here

And the %s will print all of those chars above until it meets a \0 .

You can validate that by:

printf("%d\n", testArray2 == testArray1 + 3);
// prints `1`(true)

---

As your question, there was no error because the a ... r \0 sequece in memory is owned by your process. Only the program is trying to access an address not owned by it, the OS will throw an error.