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I was making some function that takes string as function argument but

// This is working fine
char string[] = "Any string";
func(string);

//This is not working
func("Any string");

Please tell me the difference

Kapil
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    If `func` tries to modify the string then passing a literal is undefined behavior. – John3136 Feb 03 '20 at 02:26
  • Yes the function modify strings – Kapil Feb 03 '20 at 02:27
  • please tell me what do you mean by "passing the literal is undefined behaviour." – Kapil Feb 03 '20 at 02:28
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    Passing the literal is not undefined behavior. But if the function attempts to change the value of the literal, then the attempt to change it is undefined, so passing a literal to the function will lead to undefined behavior. – William Pursell Feb 03 '20 at 04:13

3 Answers3

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What is the difference in using variable as function argument and in using directly string as function argument?

char string[] = "Any string";
func(string);
// VS.
func("Any string");

string[] is a character array initialized with the size and contents of "Any string".

"Any string" is a string literal.

If func() is func(const char *s), then no difference as func() simply is reading the string.

If func() is func(char *s), yet does not modify the string pointed to by s, then then no difference as func() simply is reading the string.

If func() is func(char *s), and attempts to modify the string pointed to by s OP later reports this, then func(string) is OK as string is a modifiable character array. Yet func("Any string") is undefined behavior (UB). @John3136 @William Pursell Code should not attempt to modify a string literal. As OP reports the first works and the 2nd is "not working", this is the prime suspect.

For certainty, the definition of func() is needed as other explanations are possible.

chux - Reinstate Monica
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As per your description, it seems that the prototype of your function 

func(char* str);

OR

func(char str[]);

Now the problem comes in the datatype of the String. When you func("Any string"); which is expecting a pointer value (as an Array also works as a pointer) you are giving it a string value which will definitely generate an error as the arguments are miss-matched.

If you still want to pass a string use this:

str = "Any String";
func(str.toCharArray());
Sumit Singh
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    The “use” prescription looks more like C++ than C. Do you want to rewrite in in C or explain how this works inC? Or you can delete this. – Jonathan Leffler Feb 03 '20 at 04:23
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ok, so.... When you declare a "text string", it's type is:

const char* const text = "text string";

Now you can copy that to a char buffer, so this is ok:

char string[] = "Any string";

because 'string' in this context is a copy of the original text (with it's own memory). This however would not be ok:

char* ptr = "Some Text";

The reason is that "Some Text" is constant, and you are asking to refer to it as some memory that can be changed. This would imply to me that your 'func' has the following prototype:

void func(char* str);

Now in the above example, you could pass 'string' to it because the types match. You can't pass "Any String" though, because it's const. If however the prototype is:

void func(const char* str);

Then there should be no issue.

robthebloke
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    The first point is not correct. In C, as explained [here](https://stackoverflow.com/a/2245983/5567382), the type of a string literal is `char []`, without any `const`. You can use it anywhere you would use a `char *`, but it's undefined behavior to modify the string. – Thomas Jager Feb 03 '20 at 03:04