C Programming Language :: Warning - comparison between pointer and integer on String


	Warning - comparison between pointer and integer on String Hi, I've written a some code to assist in my understanding of strings using some of Cs built in character handling functions but I am not sure why I'm getting the following error. I hope someone can explain the errors of my ways. /* Looking for special chars in testlist / #include <stdio.h> #include <string.h> #include <ctype.h> int main () { int i; char test2[10] = "ab c1% 4!."; for (i=0; i <= 10; i++) { if ((ispunct((int) test2[i])) && (test2[i] != ".")) { printf("i= %d is punct\n",i); } } return 0; } When I use gcc -Wall -Werror to compile I get the following warning: searchstring_ispunct.c: In function `main': searchstring_ispunct.c:14: warning: comparison between pointer and integer I don't understand why the compiler is seeing the (test2[i] !=".") as an integer. Could someone possibly explain why this is the case please ? Thank you Jay J.Broe@gmail.com wrote: > Hi, I've written a some code to assist in my understanding of strings > using some of Cs built in character handling functions but I am not > sure why I'm getting the following error. > I hope someone can explain the errors of my ways. (fx:snip) > { > int i; > char test2[10] = "ab c1% 4!."; > for (i=0; i <= 10; i++) > { > if ((ispunct((int) test2[i])) && (test2[i] != ".")) (fx:snip) > I don't understand why the compiler is seeing the (test2[i] !=".") as > an integer. `test2[i]` has type `char`, which promotes to `int` in a relational expression. `"."` is a /string/, not a character - it's a literal with type `char[2]`, which decays into pointer-to-char. Your comparing the integer (value of) `test2[i]` with the pointer (to the first element of an anonynmous char array with first element '.' and second 0) ".". BOOM. -- Took Me Ages To See It! Hedgehog "The path to the web becomes deeper and wider" - October Project - Hide quoted text - J.Broe@gmail.com wrote: > Hi, I've written a some code to assist in my understanding of strings > using some of Cs built in character handling functions but I am not > sure why I'm getting the following error. > I hope someone can explain the errors of my ways. > / Looking for special chars in testlist / > #include <stdio.h> > #include <string.h> > #include <ctype.h> > int main () > { > int i; > char test2[10] = "ab c1% 4!."; > for (i=0; i <= 10; i++) > { > if ((ispunct((int) test2[i])) && (test2[i] != ".")) > { > printf("i= %d is punct\n",i); > } > } > return 0; > } > When I use gcc -Wall -Werror to compile I get the following warning: > searchstring_ispunct.c: In function `main': > searchstring_ispunct.c:14: warning: comparison between pointer and > integer > I don't understand why the compiler is seeing the (test2[i] !=".") as > an integer. test2[i] is a char, A char is an integer. "." is a string literal. It is not an integer. You are comparing a pointer (the address of the string literal) to an integer. You wanted (test2[i] != '.') since '.' _is_ a integer. On 26 May 2007 01:26:13 -0700, J.Broe@gmail.com wrote: >Hi, I've written a some code to assist in my understanding of strings >using some of Cs built in character handling functions but I am not >sure why I'm getting the following error. >I hope someone can explain the errors of my ways. It's important to note that C has no string type. The string.h functions assume '\0' terminated arrays of char. >/ Looking for special chars in testlist / >#include <stdio.h> >#include <string.h> >#include <ctype.h> >int main () >{ > int i; > char test2[10] = "ab c1% 4!."; too short, your string literal actually is "ab c1% 4!.\0" char test2[11] = "ab c1% 4!."; better: char test2[] = "ab c1% 4!."; > for (i=0; i <= 10; i++) the test2 index runs from 0 - 10, test2[10] == '\0', therfore for (i=0; i < 10; i++) prefer sizeof (for arrays) or strlen (for char) to hardcoded numbers. > { > if ((ispunct((int) test2[i])) && (test2[i] != ".")) // see other postings > { > printf("i= %d is punct\n",i); > } > } > return 0; >} -- Roland Pibinger "The best software is simple, elegant, and full of drama" - Grady Booch On May 26, 8:51 pm, rpbg@yahoo.com (Roland Pibinger) wrote: - Hide quoted text - > On 26 May 2007 01:26:13 -0700, J.Broe@gmail.com wrote: > >Hi, I've written a some code to assist in my understanding of strings > >using some of Cs built in character handling functions but I am not > >sure why I'm getting the following error. > >I hope someone can explain the errors of my ways. > It's important to note that C has no string type. The string.h > functions assume '\0' terminated arrays of char. > >/* Looking for special chars in testlist / > >#include <stdio.h> > >#include <string.h> > >#include <ctype.h> > >int main () > >{ > > int i; > > char test2[10] = "ab c1% 4!."; > too short, your string literal actually is "ab c1% 4!.\0" > char test2[11] = "ab c1% 4!."; > better: > char test2[] = "ab c1% 4!."; > > for (i=0; i <= 10; i++) > the test2 index runs from 0 - 10, test2[10] == '\0', therfore > for (i=0; i < 10; i++) > prefer sizeof (for arrays) or strlen (for char) to hardcoded numbers.> { > > if ((ispunct((int) test2[i])) && (test2[i] != ".")) > // see other postings > > { > > printf("i= %d is punct\n",i); > > } > > } > > return 0; > >} > -- > Roland Pibinger > "The best software is simple, elegant, and full of drama" - Grady Booch Ahhh I see. " " are for strings and ' ' are for chars. I was curious as I can't see the end of array what would be at the end but you've answered my question. However does that mean in a multi lined text file like this: abced 12345 would look like: abcde\0\n 12345\0\n Is this right ? Jay J.Broe@gmail.com said: <snip> > However does that mean in a multi lined text file like this: > abced > 12345 > would look like: > abcde\0\n > 12345\0\n > Is this right ? No. Text files do not contain null characters (for any reasonable definition of "text file"). When you read a line from a file, you know you've reached the end of the line when you hit the newline character. If you're reading data from a text file into an array that you wish to treat as a string, it's your job to bang a null terminator on the end (or use a function that does it for you, such as fgets). -- Richard Heathfield "Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk email: rjh at the above domain, - www.