C Language, pointers in scanf and printf

pointers in scanf and printf

Author	Message
Pavel Despo #1 / 11	pointers in scanf and printf i'm using the newest version of the djgpp compiler (gcc ported to windows). the problem is this...when i do int main(void) { char name; scanf("%s",name); printf("%s\n",name); Quote: } it doesn't work. it keeps giving me a gpf, but, int main(void) { char name; scanf("%s",&name); printf("%s\n",&name); Quote: } does work. when &name is used, isn't the address of the pointer being passed? it seems to me that we would just want to pass the pointer, the the functions can just loop through and increment the pointer to access each element. even more baffeling (to me anyway) is that the second version only works if 7 or fewer characters used. when you try it with 8 characters, it gives me a gpf. am i overlooking something? is this just a stupid mistake? please help!!! thanks.
Tue, 12 Mar 2002 03:00:00 GMT

Richard Heathfiel #2 / 11	pointers in scanf and printf Quote: > i'm using the newest version of the djgpp compiler (gcc ported to windows). > the problem is this...when i do > int main(void) > { > char name; This defines a pointer to char. It doesn't define where that pointer is pointing, and it doesn't allocate any memory. Quote: > scanf("%s",name); So this statement overwrites some random lump of memory. Bad news. Quote: > printf("%s\n",name); > } > it doesn't work. it keeps giving me a gpf, but, > int main(void) > { > char name; > scanf("%s",&name); > printf("%s\n",&name); > } > does work. Purest fluke. The code is no more correct than in the first program. Slightly less correct, if anything. when &name is used, isn't the address of the pointer being Quote: > passed? Yes. Quote: > it seems to me that we would just want to pass the pointer, Yes. Quote: > the the > functions can just loop through and increment the pointer to access each > element. Yes. Quote: > even more baffeling (to me anyway) is that the second version only > works if 7 or fewer characters used. when you try it with 8 characters, it > gives me a gpf. am i overlooking something? is this just a stupid mistake? Yes. :-) Quote: > please help!!! thanks. Okay. Reserve some space for your string. Here's an example that uses a more robust input function as well as correctly using memory: #include <stdio.h> int main(void) { char name[64]; /* Most people's names are shorter than 64. */ printf("Please type in your name: "); fflush(stdout); if(fgets(name, sizeof name, stdin) != NULL) { printf("You typed [%s]\n", name); } else { printf("The man with no name? I've always wanted to meet you.\n"); } return 0; Quote: } HTH -- Richard Heathfield "Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
Tue, 12 Mar 2002 03:00:00 GMT

Sreenivas Talasil #3 / 11	pointers in scanf and printf I assume gpf is "general protection fault"(??) and occurs if there's some memory usage violation. That's obvious because you are using char* name without allocating an memory space for the variable name. As a test try using this line name = malloc( 100 * sizeof(char)); after you declare name. Note: this is just for testing. YOU have to decide what the size of the buffer would be. NOTE2: As others are bound to point out - scanf is evil... Cheers Sreenivas Quote: > i'm using the newest version of the djgpp compiler (gcc ported to windows). > the problem is this...when i do > int main(void) > { > char name; > scanf("%s",name); > printf("%s\n",name); > } > it doesn't work. it keeps giving me a gpf, but, > int main(void) > { > char name; > scanf("%s",&name); > printf("%s\n",&name); > }
Tue, 12 Mar 2002 03:00:00 GMT

Erik Max Franci #4 / 11	pointers in scanf and printf Quote: > i'm using the newest version of the djgpp compiler (gcc ported to > windows). > the problem is this...when i do > int main(void) > { > char name; > scanf("%s",name); > printf("%s\n",name); > } > it doesn't work. it keeps giving me a gpf, ... That's because name is not initialized. It's a pointer that points nowhere, and then you try to print what it points to (as a string) and read in what it points to (as a string). Read up on pointers and strings in your C book. Quote: > ... but, > int main(void) > { > char name; > scanf("%s",&name); > printf("%s\n",&name); > } > does work. It's a coincidence that the second just doesn't happen to crash. It's also engaging in undefined behavior. -- Alcyone Systems \| irc maxxon (efnet) \| web http://www.alcyone.com/max/ San Jose, CA \| languages en, eo \| icbm 37 20 07 N 121 53 38 W USA \| Fri 1999 Sep 24 (48%/949) \| &tSftDotIotE __ / \ Love's the only reason why we all keep living \__/ Oleta Adams
Tue, 12 Mar 2002 03:00:00 GMT

Greg Mart #5 / 11	pointers in scanf and printf On Fri, 24 Sep 1999 14:33:05 -0400, Sreenivas Talasila Quote: >I assume gpf is "general protection fault"(??) and occurs if there's some >memory usage violation. >That's obvious because you are using char* name without allocating >an memory space for the variable name. >As a test try using this line >name = malloc( 100 * sizeof(char)); >after you declare name. >Note: this is just for testing. YOU have to decide what the size of the buffer >would >be. >NOTE2: As others are bound to point out - scanf is evil... So is malloc if you don't test for failure. if(name == NULL) { printf("out of memory\n"); exit(EXIT_FAILURE); Quote: } /* after your finished with name */ free(name); Regards, Greg Martin.
Tue, 12 Mar 2002 03:00:00 GMT

Lawrence Kir #6 / 11	pointers in scanf and printf Quote: >> i'm using the newest version of the djgpp compiler (gcc ported to >windows). >> the problem is this...when i do >> int main(void) >> { >> char name; >This defines a pointer to char. It doesn't define where that pointer is >pointing, and it doesn't allocate any memory. It allocates memory for the pointer itself but, say you say, nothing for it to point at. Quote: >> scanf("%s",name); >So this statement overwrites some random lump of memory. Bad news. As the most basic level it is reading the value of name which at this point is an uninitialised variable. Reading the value of an uninitialised variable is always an error, whether it be a pointer or not. Quote: >> printf("%s\n",name); >> } >> it doesn't work. it keeps giving me a gpf, but, >> int main(void) >> { >> char name; >> scanf("%s",&name); >> printf("%s\n",&name); >> } >> does work. >Purest fluke. The code is no more correct than in the first program. >Slightly less correct, if anything. Well, the only things that might be viewed as less correct than undefined behaviour are syntax errors and constraint violations and this is neither of those. There is more chance of this working than the first version on platforms that use the same representation for all data pointers and where the input string is small: at least the address sof a valid object is being passed and the string is written into that. - Hide quoted text - - Show quoted text - Quote: > when &name is used, isn't the address of the pointer being >> passed? >Yes. >> it seems to me that we would just want to pass the pointer, >Yes. >> the the >> functions can just loop through and increment the pointer to access each >> element. >Yes. >> even more baffeling (to me anyway) is that the second version only >> works if 7 or fewer characters used. when you try it with 8 characters, >it >> gives me a gpf. am i overlooking something? is this just a stupid >mistake? >Yes. :-) >> please help!!! thanks. >Okay. Reserve some space for your string. Here's an example that uses a >more robust input function as well as correctly using memory: >#include <stdio.h> >int main(void) >{ > char name[64]; /* Most people's names are shorter than 64. */ > printf("Please type in your name: "); > fflush(stdout); > if(fgets(name, sizeof name, stdin) != NULL) > { > printf("You typed [%s]\n", name); This will produce odd output for "short" names since name will contain a new-line character. Quote: > } > else > { > printf("The man with no name? I've always wanted to meet you.\n"); > } > return 0; >} -- ----------------------------------------- -----------------------------------------
Wed, 13 Mar 2002 03:00:00 GMT

Ben Pfaf #7 / 11	pointers in scanf and printf Quote: > How does Lawrence /do/ this? I sometimes wonder whether we should just > forward every question to Lawrence rather than bothering to have a go > ourselves. :-) I suspect that Lawrence isn't an actual person. In fact, he's a whole bunch of people who work together to produce answers and check them them over for correctness. The results are edited by a single person to give consistent style, then sent through for proofreading again and finally posted. The occasional typo is inserted automatically by the posting software just to throw us off and make us believe that "he"'s fallible like the rest of us. -- "I should killfile you where you stand, worthless human." --Kaz
Thu, 14 Mar 2002 03:00:00 GMT

John Winte #8 / 11	pointers in scanf and printf Quote: >> How does Lawrence /do/ this? I sometimes wonder whether we should just >> forward every question to Lawrence rather than bothering to have a go >> ourselves. :-) >I suspect that Lawrence isn't an actual person. In fact, he's a whole >bunch of people who work together to produce answers and check them >them over for correctness. Looking for things like repeated words you mean? John -- John Winters. Wallingford, Oxon, England. The Linux Emporium - a source for Linux CDs in the UK See http://www.linuxemporium.co.uk/
Fri, 15 Mar 2002 03:00:00 GMT

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