C language problem, help, thanks!

Read a character, then output it to the screen, and then judge that the character is not '? If you don't read it again, stop it.

The result is: right?

That's all about typing in any character and outputting the original character! If you don't touch the character ' during output? 'then execute while() until there is'? 'Appear, output and end the program!'

On the image you can see that the characters I typed don't end until the end!

Problem 1: The first is to figure out the compilation process when defining.

int n = 2 defines an integer variable n, assigns a value to 2p = &n, defines an integer pointer p, points to the address of variable n.

q = p；Define an integer pointer q and assign p to it, i.e. assign the address of n to it.

ABCD is just a simple assignment operation, of course, all correct, mainly to understand their meaning.

Q pointer assignment is to assign the address to pb*p = *q data assignment, which assigns the data that q now points to the address to the data that p points to.

q Data assignment, assign the value of q to the address to the variable n= &n pointer, and assign the address of the variable n to the pointer pQuestion 2: (strlen(s), you made a mistake)."Escape symbols, one is"\"meaning;

x"an escape symbol, which may indicate an unsigned hexadecimal;

The escaped symbol belongs to this string, but does not occupy the length of the string. Therefore, the string is"\41xyz", the true length is 5, and the output is 5.

In the first question, all four are valid, p points to n, p is assigned to q, and p and q are both pointers to n.

The second problem, compilation error, strlen accepts the first address of the string, if you really want to learn a little better, you can debug it more.

There are a lot of questions, and my humble opinion is as follows:

1. We know that the name of an array is the first address of the array, and the one is a pointer, so a is a pointer to a two-dimensional array. (We might as well write a[0], a[1] as x, y.) Then the array that a points to is x[3][4], y[3][4])).

2. Because the identifier in front of the number of elements of an array can be regarded as the array name of the array, then a[0] here is the array name of the two-dimensional array (equivalent to the x mentioned above). And just now we see that we can think of a[0] as the x in x[3][4], then we know that x actually points to a one-dimensional array of size 4, then a[0] points to a one-dimensional array of size 4.

In addition, a is the array name of the three-dimensional array, so it represents the first address of the three-dimensional array (that is, the first address of 0 rows), that is, &a[0][0], and a[0] we just mentioned, it is equivalent to the first address of the two-dimensional array, that is, the first address of 0 rows and 0 columns, so it is &a[0][0][0]. And a[0][0] is equivalent to the first address of a one-dimensional array, that is, the first address of 0 rows, 0 columns, and 0 verticals, so it is also &a[0][0][0].