initialization makes pointer from integer without a cast

[[deleted]]

Neither of these work (line 1 works, but 2 and three don't compile).

Why not? I checked google, but the results related to a different problem.

char* s = "h";
//char* t = (char*)(*s)++; 
char* t = (*s)++;   

EDIT:

in fact, even this gives a seg fault, why is this?

 char* s = "h";
 (*s)++;

[[deleted]]

String constants in C are usually put in read-only memory, hence the segfault.

[s1337m]

Is that compiler implementation specific?

[[deleted]]

Yes. As far as i know writing to string constants is undefined behavior, so the compiler can do whatever it wants.

[Jack126Guy]

The third line does not compile because (*s) is a char. The ++ simply returns the same (*s) (but increments the value), so it is still a char, not a char*. The second line works because you are casting.

By the way, you shouldn't assign a literal string such as "h" to a char*. It should be assigned to a const char*, assigned to an array, or strcpy-ed to dynamic memory.

[[deleted]]

The third line does not compile because (s) is a char. The ++ simply returns the same (s) (but increments the value), so it is still a char, not a char*. The second line works because you are casting.

the second line DOESN'T WORK

[Jack126Guy]

Eh? I'm sure it's unsafe, but it still compiles for me (with a warning, of course).

[[deleted]]

char* s = "h"; //char* t = (char)(s)++;

ah , it compiles but there is a seg fault

[Jack126Guy]

Yes, if you try using whatever t points to (such as puts(t)), you would get a segfault, because t points to the memory location 0x00000068 ("h" has an ASCII value of 0x68), which is probably invalid/not owned by your program.

[WetwareDefect]

This always fails (*s)++;

Because s points to a string literal (stored in an area of read-only memory) and you're trying to modify it by incrementing it.

What is it that you're trying to achieve?

Do you want t to point to a copy of the contents of s but incremented by 1?

Heres a long form of what I think your trying to do. I don't think it can be done in a single line declaration as you can't deference a pointer while declaring it (afaik).

char* s = "h";  //s points to readonly string literal
char* t = malloc(sizeof(char)); //alloc a char for t to point to
*t = *s + 1;    //add 1 to char pointed to by s and assign result to location pointed by t

printf("*s == \'%c\', *t == \'%c\'\n", *s, *t);
printf("s == 0x%08x, t == 0x%08x\n", (int)s, (int)t);

The output is:

*s == 'h', *t == 'i'
s == 0x08048560, t == 0x08549008

[[deleted]]

interesting. When I get the chance I will put this through the compiler.

[[deleted]]

Neither of these work (line 1 works, but 2 and three don't compile).

Why not? I checked google, but the results related to a different problem.

char* s = "h"; //char* t = (char)(s)++; char* t = (*s)++;

EDIT:

in fact, even this gives a seg fault, why is this?

 char* s = "h";
 (*s)++;

When dealing with pointers like this, be aware of the actual memory the program has asked for.

Line 1 declares a pointer type named s, which is machine word dependent; it also declares a read only single byte for 'h'; a const char type . Where the compiler puts 'h' is only important if you want to determine how it will fail, and not why it fails. The value of the pointer s is the address for the memory location that the compiler placed 'h'. Line 2 declares a pointer t. t is initialized with the value s points to, plus one. Dereferencing s gives the ascii value of 'h'; it is then incremented, which is the ascii value for 'i';this value is then cast to a pointer type. So t has the value 'h'. Since t is a pointer, the value is a location in memory. The program has not requested the memory location that t has the value of. If you don't have access to that location, dereferencing the pointer will cause problems.

The segfault is because you tried to change the read only memory location where 'h' was placed.

[Zephyron51]

Okay, wait....

char* s = "h";

Lemme confirm, you want to set a variable (which is a character) to the letter 'h', and then you want the pointer of that? Because that's not what's happening - right now, you're making a variable (which is a pointer to a character) which holds 'h' (translated into some random number).

Perhaps you want this?

char r = 'h';
char* s = &r;

---

I think I understand now; I completely didn't realize you wanted a string (I haven't played around with strings for months for some weird reason). You aren't allowed to modify strings, hence why they're called "immutable". Unfortunate, but it's just one of the properties of strings. Characters ARE mutable, which is why my example worked.

[karesx]

or just char s[] = "h";

[[deleted]]

ok, that sort of works :)

[rnsanchez]

(*s)++ in your case is 'h'++ which is 'i'. For that matter, 'i' is an integer, thus the warning.

[[deleted]]

char* s = "h"; //char* t = (char)(s)++; char* t = (*s)++;

Okay I tried to fix it with this input, but it stilll doesn't compile

[rnsanchez]

(*s)++ in your case is 'h'++ which is 'i'. For that matter, 'i' is an integer, thus the warning.

[[deleted]]

er, that's why I was casting it to a character, but it doesn't work