String initializer and read only section

Question

Suppose that I have an array(local to a function) and a pointer

char a[]="aesdf" and char *b="asdf"

My question is whether in the former case the string literal "aesdf" is stored in read only section and then copied on to the local array or is it similar to

char a[]={'a','e','s','d','f','\0'}; ?

I think that in this case the characters are directly created on the stack but in the earlier case (char a[]="aesdf") the characters are copied from the read only section to the local array.

Will `"aesdf" exist for the entire life of the executable?

Answer 1

Both a[] ="aesdf" and char a[]={'a','e','s','d','f','\0'} will be stored in function's run time stack and memory will be released when function returns. but for char* b= "asdf" asdf is stored in readonly section and is referred from there.

Answer 2

char a[] = "aesdf";
char a[] = {'a','e','s','d','f','\0'};

These two lines of code have the same effect. A compiler may choose to implement them the same way, or it may choose to implement them differently.

From the standpoint of a programmer writing code in C, it shouldn't really matter. You can use either and be sure that you will end up with a six element array of char initialized to the specified contents.

Will "aesdf" exist for the entire life of the executable?

Semantically, yes. String literals are char arrays that have static storage duration. An object that has static storage duration has a lifetime of the execution of the program: it is initialized before the program starts, and exists until the program terminates.

However, this doesn't matter at all in your program. That string literal is used to initialize array a. Since you do not obtain a pointer to the string literal itself, it doesn't matter what its actual lifetime of that string literal is or how it is actually stored. The compiler may do whatever it sees fit to do, so long as the array a is correctly initialized.

Answer 3

From the abstract and formal point of view, each string literal is an independent nameless object with static storage duration. This means, that the initialization char a[] = "aesdf" formally creates literal object "aesdf" and then uses it to initialize the independent array a, i.e. it is not equivalent to char *a = "aesdf", where a pointer is made to point to the string literal directly.

However, since string literals are nameless objects, in the char a[] = "aesdf" variant there's no way to access the independent "aesdf" object before or after the initialization. This means that there's no way for you to "detect" whether this object actually existed. The existence (or non-existence) of that object cannot affect the observable behavior of the program. For this reason, the implementation has all the freedom to eliminate the independent "aesdf" object and initialize the a array in any other way that leads to the expected correct result, i.e. as char a[] = { 'a', 'e', 's', 'd', 'f', '\0' } or as char a[] = { 'a', 'e', "sdf" } or as something else.

Answer 4

First:

char a[]="aesdf";

Assuming this is an automatic local variable, it will allocate 6 bytes on the stack and initialize them with the given characters. How it does this (whether by memcpy from a string literal or loading a byte at a time with inline store instructions, or some other way) is completely implementation-defined. Note that initialization must happen every time the variable comes into scope, so if it's not going to change, this is a very wasteful construct to use.

If this is a static/global variable, it will produce a 6-byte char array with a unique address/storage whose initial contents are the given characters, and which is writable.

Next:

char *b="asdf";

This initializes the pointer b to point to a string literal "asdf", which might or might not share storage with other string literals, and which produces undefined behavior if you write to it.