Programming Forums > Application Development > C

conbrio · Apr 16th, 2005, 5:10 AM

okay, i'm new to C and going crazy trying to understand this little problem.

in the code below, example A doesn't work, while B does. why is this so?

i thought that declaring strings by "char *blah" and "char blah[]" would essentially be the same since arrays are just pointers anyway. am i wrong?

the reason i ask this is because i'm writing a function that needs to use strings which aren't initialized with a value and hence need to have the array size automatically known ("char *blah;" works for this while "char blah[];" gives an array size unknown error).

after all this, i then want the strings to be passed to strcat() for concatenation.

i really hope someone can help here.. thanks.

(Toggle Plain Text)

// EXAMPLE A
#include <stdio.h>
#include <string.h>



main()
{
   char *yr="1985";
   char *car=" toyota";
   strcat(yr, car);
   printf("%s", yr);
}



// EXAMPLE B
#include <stdio.h>
#include <string.h>


main()
{
   char yr[]="1985";
   char car[]=" toyota";
   strcat(yr, car);
   printf("%s", yr);
}

BaroN NighT · Apr 16th, 2005, 7:12 AM

Pointers points to an address of a variable, thus example A will not work because its 1985 and toyota is not an address.
However, example B works because you are declaring a char string to yr[] and car[]. Both 1985 and toyota in this example fills the array.
Arrays are different from pointers, for yr[], yr works like a pointer but it only points to the address of the first element in the string.

Benoit · Apr 16th, 2005, 11:55 AM

Arrays are just pointers in disguise

*(ara+2) is the same thing as ara[2]

Example A didn't work because *yr didn't have enough space to hold *car

Eggbert · Apr 16th, 2005, 2:23 PM

>>would essentially be the same since arrays are just pointers anyway
Arrays are not pointers and pointers are not arrays. An array name is converted to a pointer to the first element in many situations, so it may seem that way, but mistaking one for the other results in subtle and dangerous errors.

>>example A doesn't work, while B does. why is this so?
Actually, both are wrong. Example A attempts to modify a string literal and example B tries to copy more data into an array than the array can hold. Here are the differences.

>>char *yr="1985";
yr is a pointer to a string literal. The string literal may be placed in read-only memory, so it is effectively constant. And attempt to modify the memory that yr points to is an error.

>>char yr[]="1985";
yr is an array that was initialized with a string literal. The size of the array-because it was omitted-is determined by the length of the string literal, including the null character. So the length of yr is 5 and it was initialized with {'1','9','8','5','\0'}. Because yr resides within your address space and is not read-only, you can modify it.

Apr 16th, 2005, 5:10 AM	#1
conbrio Newbie Join Date: Apr 2005 Posts: 10 Rep Power: 0	pointers strings, and strcat() okay, i'm new to C and going crazy trying to understand this little problem. in the code below, example A doesn't work, while B does. why is this so? i thought that declaring strings by "char blah" and "char blah[]" would essentially be the same since arrays are just pointers anyway. am i wrong? the reason i ask this is because i'm writing a function that needs to use strings which aren't initialized with a value and hence need to have the array size automatically known ("char blah;" works for this while "char blah[];" gives an array size unknown error). after all this, i then want the strings to be passed to strcat() for concatenation. i really hope someone can help here.. thanks. (Toggle Plain Text) // EXAMPLE A #include <stdio.h> #include <string.h> main() { char yr="1985"; char car=" toyota"; strcat(yr, car); printf("%s", yr); } // EXAMPLE B #include <stdio.h> #include <string.h> main() { char yr[]="1985"; char car[]=" toyota"; strcat(yr, car); printf("%s", yr); } // EXAMPLE A #include <stdio.h> #include <string.h> main() { char yr="1985"; char car=" toyota"; strcat(yr, car); printf("%s", yr); } // EXAMPLE B #include <stdio.h> #include <string.h> main() { char yr[]="1985"; char car[]=" toyota"; strcat(yr, car); printf("%s", yr); }

Apr 16th, 2005, 11:55 AM	#3
Benoit Expert Programmer Join Date: Sep 2004 Location: Ontario, Canada Posts: 579 Rep Power: 5	Arrays are just pointers in disguise (ara+2) is the same thing as ara[2] Example A didn't work because yr didn't have enough space to hold *car __________________ Johnny was a chemist's son but Johnny is no more, for what Johnny thought was H2O was H2SO4

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Apr 16th, 2005, 7:12 AM	#2
BaroN NighT Programmer Join Date: Mar 2005 Location: USA Posts: 60 Rep Power: 4	Pointers points to an address of a variable, thus example A will not work because its 1985 and toyota is not an address. However, example B works because you are declaring a char string to yr[] and car[]. Both 1985 and toyota in this example fills the array. Arrays are different from pointers, for yr[], yr works like a pointer but it only points to the address of the first element in the string.

Apr 16th, 2005, 2:23 PM	#4
Eggbert Professional Programmer Join Date: Nov 2004 Posts: 250 Rep Power: 5	>>would essentially be the same since arrays are just pointers anyway Arrays are not pointers and pointers are not arrays. An array name is converted to a pointer to the first element in many situations, so it may seem that way, but mistaking one for the other results in subtle and dangerous errors. >>example A doesn't work, while B does. why is this so? Actually, both are wrong. Example A attempts to modify a string literal and example B tries to copy more data into an array than the array can hold. Here are the differences. >>char *yr="1985"; yr is a pointer to a string literal. The string literal may be placed in read-only memory, so it is effectively constant. And attempt to modify the memory that yr points to is an error. >>char yr[]="1985"; yr is an array that was initialized with a string literal. The size of the array-because it was omitted-is determined by the length of the string literal, including the null character. So the length of yr is 5 and it was initialized with {'1','9','8','5','\0'}. Because yr resides within your address space and is not read-only, you can modify it.

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Programming Forums > Application Development > C

pointers strings, and strcat()