Ben Pfaff <bl*@cs.stanford.edu> wrote in message news:<87************@pfaff.stanford.edu>...
vi**********@hydbad.tcs.co.in (Vishal Ladha) writes:
Code1 :
=====
char *ptr = "hello";
if (ptr == "hello")
{
printf("true\n");
}
else
{
printf ("false\n");
} [...] Now, I know that Code1 is a pathetic way of programming and it
will result in a segmentation fault in some memory operation(strcpy etc).
It's not useful, but there's nothing wrong with it from a
standards perspective. It certainly doesn't invoke undefined
behavior.
Maybe you mean that if you try to strcpy() over ptr as assigned
above, you'll get undefined behavior. That's true. But it's
really quite unclear to me what you actually mean.
I think I have got the answer from Darrell Grainger. Here it is
================================================== ==============
Hi !
I have been experimenting with char * for a while now. I have two pieces of
code :
Code1 :
=====
char *ptr = "hello";
This line of code creates two things. The first thing is a variable ptr.
The second thing is the string literal "hello". It then assigned the
address of the string literal to the variable ptr. For example, if the
compiler puts the string literal "hello" in memory location 5000 then ptr
equals 5000.
if (ptr == "hello")
The compiler could create a second string literal "hello" but a good
compiler will know that a constant string literal "hello" already exists
and it will use that same string for this comparison. This means that the
string literal here *CAN* be the same memory location as the one above.
In the case of your compiler, they are the same string. Therefore, the
comparison will be true. It is equally possible for the comparison to be
false. For example, the above assignment makes "hello" a constant string
literal. Change it to:
char ptr[] = "hello";
and you *MIGHT* find this if statement is false.
{
printf("true\n");
}
else
{
printf ("false\n");
}
Code2:
=====
char *ptr = (char *) malloc(6* sizeof(char));
strcpy(ptr,"hello")
In this case, ptr is being assigned the address of a writable memory
location. The string literal "hello" *MIGHT* be read only. In all cases,
the memory location of the malloc memory and the memory location of the
string literal "hello" will be different.
if (ptr == "hello")
Since ptr holds the address of the malloc memory and that is a different
memory location then the string literal "hello", this will be false.
{
printf("true\n");
}
else
{
printf ("false\n");
}
Code1 prints true while Code2 prints false.
Yes. Code1 can print true or false depending on the compiler. Code2 will
always print false.
Now, I know that Code1 is a pathetic way of programming and it
will result in a segmentation fault in some memory operation(strcpy etc).
It depends on the compiler. If your compiler places string literals, like
"hello", in read only memory locations then attempting to alter it, via
ptr, will result in a segmentation fault. If your compiler places string
literals in writable memory then it will not cause a segmenttation fault.
But I have 2 questions regarding this.
1. Why are the the answers different in Code1 and Code2. I had expected
the answers in both the cases to be false.
See above.
2. What is the difference between == operator and stcmp ?.
Why doesn't == work for strings (ASCII and binary comparison ????)
A string is a concept. There is actually no atomic data type called a
string in C language. A string is REALLY an array of char with a '\0'
character marking the end of the useful array. The == operator cannot be
used to compare arrays. When you pass the name of an array to a function
the receiving function actually receives the address of the first element.
To be consistent, the == operator works the same way. Thus if I have:
char s1[] = "hello";
char s2[] = "hello";
if(s1 == s2)
puts("Will never get here.");
else
puts("This will always print");
It is the same as:
char s1[] = "hello";
char s2[] = "hello";
if(&s1[0] == &s2[0])
puts("Will never get here.");
else
puts("This will always print");
The address of the start of s1 will not be the same as the address of s2.
The contents are the same but the memory locations are different.
================================================== =====================
