String in function parameter

若如初见. 提交于 2019-11-27 15:04:40

问题


int main()
{
        char *x = "HelloWorld";
        char y[] = "HelloWorld";

        x[0] = 'Z';
        //y[0] = 'M';

        return 0;
}

In the above program, HelloWorld will be in read-only section(i.e string table). x will be pointing to that read-only section, so trying to modify that values will be undefined behavior.

But y will be allocated in stack and HelloWorld will be copied to that memory. so modifying y will works fine. String literals: pointer vs. char array

Here is my Question:

In the following program, both char *arr and char arr[] causes segmentation fault if the content is modified.

void function(char arr[])
//void function(char *arr)
{
   arr[0] = 'X';
}        
int main()
{
   function("MyString");    
   return 0;
}
  1. How it differs in the function parameter context?
  2. No memory will be allocated for function parameters??

Please share your knowledge.


回答1:


Inside the function parameter list, char arr[] is absolutely equivalent to char *arr, so the pair of definitions and the pair of declarations are equivalent.

void function(char arr[]) { ... }
void function(char *arr)  { ... }

void function(char arr[]);
void function(char *arr);

The issue is the calling context. You provided a string literal to the function; string literals may not be modified; your function attempted to modify the string literal it was given; your program invoked undefined behaviour and crashed. All completely kosher.

Treat string literals as if they were static const char literal[] = "string literal"; and do not attempt to modify them.




回答2:


function("MyString");

is similar to

char *s = "MyString";
function(s);

"MyString" is in both cases a string literal and in both cases the string is unmodifiable.

function("MyString");

passes the address of a string literal to function as an argument.




回答3:


char *arr; above statement implies that arr is a character pointer and it can point to either one character or strings of character

& char arr[]; above statement implies that arr is strings of character and can store as many characters as possible or even one but will always count on '\0' character hence making it a string ( e.g. char arr[]= "a" is similar to char arr[]={'a','\0'} )

But when used as parameters in called function, the string passed is stored character by character in formal arguments making no difference.



来源:https://stackoverflow.com/questions/17111140/string-in-function-parameter

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