Pass string literal as char* via argument

Question

I made a function which change string, see the following code.

void Test(char* str, char c) {
    str[1] = c;
}

int main(){
    Test("Hi", '2');
}

I notice it made some run time error. I know how to prevent the error.

char buff[3] = "Hi";
Test(buff,'2');

but I don't know why the first example made run time error. I guess, if I pass string directly, it becomes const char. Does anyone explain what happened exactly?

ps. what if I use char* str = "hi", then pass it into the argument?

char* buff = "Hi";
Test(buff,'2');

like this. Can I modify buff?

This isn't legal C++ any more (as in shouldn't compile) and has never been right. — Qaz
– Qaz, Commented Jan 15, 2014 at 8:07
The last one being new, I presume it's supposed to be char *buff, but still no. Using a pointer to non-constant data to point to constant data will never work regardless of how you try to get around it. — Qaz
– Qaz, Commented Jan 15, 2014 at 8:12

Kiril Kirov · Accepted Answer · 2014-01-15 08:15:14Z

3

Because "Hi" is string literal and it's not allowed to be modified, they are read-only (the type of string literal is const char[n]).

Modifying it is undefined behavior.

Regarding your edit: char* str = "hi" is invalid, it should be const char* str = "hi". Which is pointer to const char. Again, modifying it is disallowed.

edited Jan 15, 2014 at 8:15

answered Jan 15, 2014 at 8:04

Kiril Kirov

38.5k26 gold badges120 silver badges195 bronze badges

Sign up to request clarification or add additional context in comments.

11 Comments

barak manos Over a year ago

True if he had declared char* buff = "Hi" (which would place buff in the read-only code section instead of in the stack).

Konrad Rudolph Over a year ago

@barak OP did (implicitly) declare it like that. This answer is correct.

Qaz Over a year ago

Note: const char[N], though.

falstro Over a year ago

@barakmanos; you need to double check the question, it's the first case which is giving him problems

barak manos Over a year ago

@Kiril Kirov: Try for yourself: When char buff[3] = "Hi", buff is placed in the stack of the calling function, or in the data-section of the program (if declared globally). Both the stack and the data-section are Read-Write sections. When 'const char buff[3] = "Hi"` or char* buff = "Hi", `buff is placed in the code-section of the program, which is a Read-Only section.

|

Siva Krishna Aleti · Accepted Answer · 2014-01-15 08:25:01Z

When you don't explicitly allocate memory for strings, compiler stores them in read-only memory. So, any modification to such strings result in run time error.

Test("Hi", '2');

Here in the above case "Hi" string is stored in read-only memory.

 char *buff = "Hi";  
 Test(buff,'2');

Here also "Hi" is stored in the read-only memory and the starting address is returned to buff character pointer, which is same as above. You can overcome such errors by allocating memory for the string and then pass that reference. Like

 char buff[3] = "Hi";  
 Test(buff,'2');

or

char *buff = (char *)malloc(SIZE);  
strcpy(buff, "Hi");
Test(buff,'2');

Please refer to this link http://www.geeksforgeeks.org/memory-layout-of-c-program/

trojanfoe · Accepted Answer · 2014-01-15 08:04:31Z

0

Often string constants are in read-only memory, causing a runtime error when you attempt to modify it.

In your second example, you put the string into a buffer on the stack, so it can be updated without error.

answered Jan 15, 2014 at 8:04

trojanfoe

123k23 gold badges219 silver badges249 bronze badges

Comments

Barmar · Accepted Answer · 2014-01-15 08:05:23Z

0

Literal strings are not modifiable. When I compile your code with GCC I get the warning:

testptr.cpp:6: warning: deprecated conversion from string constant to 'char*'

answered Jan 15, 2014 at 8:05

Barmar

789k57 gold badges555 silver badges669 bronze badges

Comments

barak manos · Accepted Answer · 2014-01-15 08:29:26Z

Runtime Error:

char* buff = "Hi"; // buff points to an address in the code-section, which is a Read-Only section
buff[1] = 'x';     // Illegal memory access violation

Compilation Error:

const char* buff = "Hi"; // This is a correct declaration, which will prevent the runtime error above
buff[1] = 'x';           // The compiler will not allow this

All Good:

char buff[] = "Hi"; // buff points to an address in the stack or the data-section, which are both Read-Write sections
buff[1] = 'x';      // Works OK

Notes:

In all cases, a string "Hi" is placed in the code-section of the program.
In the last example, the contents of that string are copied into the buff array.
In the last example, the buff array is located in the stack if buff is a non-static local variable, and in the data-section of the program otherwise.

Collectives™ on Stack Overflow

Pass string literal as char* via argument

5 Answers 5

11 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

11 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related