This is part of an assignment that I'm supposed to submit, this program is supposed to take 2 inputs, a name containing 4 letters and a city containing 5 letters then sub them in that sentence. The problem is that the first word is not being output properly, like if I type john all I get is 'r' instead of the word. I'm new to this and there might be other mistakes, but I wanna fix this one. Help greatly appreciated. :l
#include <stdio.h>
int main(int argc, char **argv)
{
char name1[4];
char city1[5];
printf("\nEnter four letter name:\n");
scanf("%s", name1);
printf("\nEnter five letter city:\n");
scanf("%s", city1);
printf("%s was afraid of the airplane, he walked from %s to Romaine.", name1, city1);
return 0;
}
When I put John and Denver it output "r was afraid of the airplane he walked from Denver to Romaine"
Aham, so this is a buffer overflow. What you should do is:
I. Allocate two reasonably long buffers. 4 and 5 just make me cry. There's a LINE_MAX macro in <limits.h> which may be useful.
II. Use a safe function that lets you specify how big your buffer is. For example, fgets() is an excellent function for inputting a line of text. Using it also has the benefit that it really always reads an entire line (if the buffer passed to it is large enough). scanf() requires some messing with the %s conversion specifier before you can make it accept whitespace and stuff.
char name[LINE_MAX];
char city[LINE_MAX];
fgets(name, sizeof name, stdin);
fgets(city, sizeof city, stdin);
#include <stdio.h>
int main(int argc, char **argv)
{
char name1[8];
char city1[8];
printf("\nEnter four letter name:\n");
scanf("%s", name1);
printf("\nEnter five letter city:\n");
scanf("%s", city1);
printf("%s was afraid of the airplane, he walked from %s to Romaine.", name1, city1);
return 0;
}
The solution: Make sure the character arrays are large enough for the input.
scanf is an unsafe function as it will fetch data from its input stream and copy it past the end of the buffer you provide if it's not large enough. In addition it will place a nul terminating character at the end of the copied string.
In your example John requires at least 5 characters while Denver at least 7 because you must account for the terminating character.
You should consider using a larger buffer for both strings, or as a better alternative, write your own function which uses a buffer to store the string and then copies it to the final destination buffer (assuming length is enough, truncating otherwise).
Take a look at this answer: How to prevent scanf causing a buffer overflow in C?, there are some common work arounds for your problem.
Related
This question already has answers here:
Why does scanf ask twice for input when there's a newline at the end of the format string?
(7 answers)
Closed 5 years ago.
I am trying to simply get a user input string. The string is saved (scanf) when the person hits enter or the string hits the NULL character. Yet, when I run my program, and type in the string, it continues input until I type \n or \0 (whichever I have in my if statement. Here's my code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MSL 30
char intersperse(char);
char widen_stars(char);
int main(int argc, char *argv[])
{
char *p, *q, *str1[MSL], *str2[MSL];
if(str1[MSL+1] != '\0'){
printf("Please enter a string of maximum 30 characters: ");
scanf("%s\n", str1[MSL]);
}
printf("%s\n", str1[MSL]);
}
This is by no means finished code, i'm just in the process of writing my program when I snagged this annoying bug. The +1 on my MSL is to make sure the NULL character is read so my string doesn't yell at me, idk if it's necessary, but it's a precaution. Thank you answerers.
There are various problems with your code, not just a simple bug. I'll try to describe some of them here.
Turn on compiler warnings. For example, if your array is size 30, there is no element at index 30 (indexes would go from 0 to 29). Your compiler could have warned you about that and other problems.
Store a string in a char array or pointer, but not in a char pointers array. When you wrote:
char *str1[MSL]
You are creating an array of pointers to char, or an array of strings. However, since you're dealing with pointers to char as strings, you'd need to allocate space for them yourself. That's another concept. You probably meant to write an array of char like this:
char str1[MSL]
char *p, *q, str1[MSL], str2[MSL]; /* in your declaration */
Enter at maximum 29 characters. If your string holds 30 char, and you need 1 to mark the null at the end, you've got 29 remaining usable chars. Or… change MSL to 31.
Consider dropping the \n from your scanf format. It will make it read all whitespace (which is what \n represents in a format) waiting for the next non-whitespace. Thus, the input somestring<enter> for example won't be sent directly to your program until the next non-whitespace or EOF.
Give scanf an address. That could be just the array name (which translates to the first element address) or a pointer if you were working with them.
scanf("%s", str1);
scanf("%s", strptr); /* strptr is a char pointer pointing
to previously allocated space */
You can limit the length of what scanf reads. To be safe:
scanf("%29s", str1);
Don't use uninitialised data in comparisons. When you wrote if(str1[MSL+1] != '\0'), what did you expect str1 to contain if you had never stored anything yet?
int
main(int argc, char **argv)
{
char *p, *q, str1[31], str2[31]; /* a lot of unused variables */
printf("Please enter a string of maximum 30 characters: ");
scanf("%30s", str1);
printf("%s\n", str1);
return 0;
}
My code looks like this:
int nameFull;
printf("What is your name?\n");
scanf("%d\n", &nameFull); \\up until here it seems to work
printf("Hello %d", nameFull);
return 0;
But my output every time I run the program is "Hello 0" no matter what I input.
Does anyone know how to fix this?
First of all scanf() doesn't emit a prompt so its not a good idea to use any trailing whitespace character in the format string like \n here , It will cause it to read and discard character until next non-whitespace character.
To read a name you can do it like :
char name[50];
scanf("%49s",name); // 49 to limit the buffer input to prevent buffer overrun , this is a security issue.
You should also check the return value of scanf to see if the operation was successful. Personally , I don't prefer using scanf() at all because of various potential problems. It takes as input only what the program author expects it to, not considering other inputs which user might accidentally input. Check out here and here. Also check the scanf() man page
A better and safer method would be use fgets(),
fgets(name,sizeof(name),stdin);
You want to read a string, but you are an integer to store the input. That's not the right approach.
A better aproach would be to use an array of characters, to store the string in it.
char nameFull[100]; // can store up to 100 characters, 99 + 1 for the null-terminator ideally
Now, you could use scanf, like this:
scanf(" %99[^\n]", nameFull);
Note that I used 99, as a guard for not overflowing your array nameFull, if the user inputs too many characters for the size of your array. I didn't use %s, which would stop at a whitespace, and you seem to want to input a full name, which is usually two words and a space in between.
An alternative would be to use fgets(), which provides more safety, like this:
fgets(nameFull, sizeof(nameFull), stdin)
It will read the whole line though and store the trailing newline, while scanf() will read a single string.
Moreover, use the string identifier to print, not the integer one (%s is for string, %d is for integers). Like this:
printf("Hello %d", nameFull);
to this:
printf("Hello %s", nameFull);
as discussed about the string format.
%s reads a string of characters.
%d reads a integer.
So, your correct code will be like following code :
#include <stdio.h>
int main(){
char nameFull[100];
printf("What is your name?\n");
scanf("%99s", nameFull); //to avoid potential buffer overflow
printf("Hello %s\n", nameFull);
return 0;
}
N.B: Check this comment for nice explanation.
Well, int stores a number, a name is not a number. A name is a set of characters (aka strings). So this program would work (no error checking and such since you are in an introductory course):
char name[1024]; // 1024 is more than enough space for a name
scanf("%s", name); // %s reads a string of characters
printf("Hello %s\n", name);
return 0;
You are trying to assign an array of character (commonly referred as string) to an integer variable.
That's not correct.
Just change your variable as such
char nameFull[1024] = {0};
And then use scanf(3) with the appropriate format specifiers for strings, which is %s
scanf("%s", nameFull);
Normally you would check for the return of scanf to know if errors occurs, and in such cases, handle them.
Anyway, I would advice you to use fgets(3) which prevents buffer overflow
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte (aq\0aq) is stored after the last character in the buffer.
My goal with this program is to incorporate the users inputs into a sort of interactive/randomized story but I'm not sure how I'm supposed to get the inputs from the users to fit between *ptrDescription, *ptrBeginning, *ptrMiddle, and *ptrEnd. Any help would be much, much appreciated!
#include <stdio.h>
#include<stdlib.h>
#include<time.h>
#include <string.h>
#include <ctype.h>
int main(void){
int i;
char name[20];
char color[20];
int age;
char sentence[1];
//array of pointers to char arrays
char *ptrDescription[]={"the painfully handsome","the one and only","who seemed much older than"};
char *ptrBeginning[]={"was blissfully ignoring","could clearly see","had no idea"};
char *ptrMiddle[]={"the huge truck","the falling meteor","the bucket of milk","the mailman","the most powerful wizard"};
char *ptrEnd[]={"that was barreling toward them.","on the horizon."};
srand(time(NULL));
printf("Enter your first name: ");
scanf("%s", &name);
printf("\nEnter your age: ");
scanf("%d", &age);
printf("\nEnter your favorite color: ");
scanf("%s", &color);
for (i = 0; i < 1; i++)
{
//strcpy(sentence,ptrDescription[rand()%3]);
//strcat(sentence," ");
//strcat(sentence,ptrBeginning[rand()%3]);
//strcat(sentence," ");
//strcat(sentence,ptrMiddle[rand()%5]);
//strcat(sentence," ");
//strcat(sentence,ptrEnd[rand()%2]);
//strcat(sentence,".");
//sentence[0]=toupper(sentence[0]);
puts(sentence);
}
getch();
return 0;
}
EDIT:
I've edited a section of my code so that directly following for (i = 0; i < 1; i++) it now looks like this:
snprintf(sentence, sizeof sentence,"%s, %s %d year old, %s %s %s %s", name, ptrDescription[rand()%3], age,ptrBeginning[rand()%3], ptrMiddle[rand()%5], ptrEnd[rand()%2]);
There are tons of strange characters after the sentence in the output, like Japanese characters and stuff. I'm not sure why they're there, though. This is what it looks like exactly:
"Enter your first name: Justin
Enter your age: 20
Justin, the arrogant 20 year old, was purposefully ignoring the most powerful wizard that was barreling toward them. 汽$0HβHζ(ﾃﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌH・(DｷHH広$0陏&・汽$0ﾀHζ(ﾃﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌﾌH WH・ H櫛H・t9HνHﾃ<"
Anyone know how I can get rid of them?
If you already have a name and an age, it's just a matter of inserting them into the correct place in sentence, right? So strcat(sentence, name) would work for name. age is a little trickier since you have to format the number first, and strcat won't do it for you. One solution would be to use sprintf(buf, "%d", age), and then concatenate buf (which is a scratch char array you would have to declare).
Any time you work with strings in C, you have to be concerned about having enough space in the target buffer. Your program can run out of space during both input and output. For the output, I would get rid of sentence altogether; since you just end up writing to stdout I would printf("%s", [part]) each part as you go along. For reading, scanf supports adding a length argument to the format string.
If you use one of the *printf functions, there are 2 things you must be careful about:
The arguments you pass are correct for the format string you use
Your buffer ends up null-terminated
Your current problem is with #1 - your format string promises 7 arguments to follow, but you only supply 6. snprintf grabs a "random" 7th value from the stack, interprets it as a char pointer, and copies whatever it finds there to sentence. You could see similar problems if your format string promised a char pointer but you placed an int in a given position. In this case the format string is a constant, so a smart compiler can validate that your format string matches the subsequent parameters. You'll want to get into the habit of taking compiler warnings seriously and not ignoring them.
The second point could be an issue if your sentence ended up bigger than your sentence buffer. If there is no room for a null-terminator, one won't be applied. You can check the return value of snprintf, or you can defensively always write a 0 to the last array position.
Here you can see my source code:
#include <stdio.h>
int main()
{
char yourname;
char yoursex;
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &yourname);
printf("Oh, hello %s! \n\n", &yourname);
printf("Are you a boy or a girl?: ");
scanf("%s", &yoursex);
printf("Nice to know you are a %s! \n\n", &yoursex);
printf("How old are you %s? I am ", &yourname);
scanf("%d", &yourage);
printf("I see you are %d, you have many years then!", &yourage);
return 0;
}
I was trying things that I didn't knew, and strangely it is not working for me. What's the problem? Also, why it needs to be %s and not %c? If I use %c instead it does not work!
Where it says:
How old are you %s? instead of putting my name, it says ''oy''
and instead of showing my age in the last line, it shows a big number.
These are the very basics of C Programming, and I strongly advise you to get a decent book - The C Programming Language by Dennis Ritchie would be a good start.
There are numerous errors in your code.
A char can contain only one character, like 'A', or 'a' or something like that. When you're scanning a name, it is going to be a group of characters, like 'E', 'd', 'd', 'y'. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.
When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c\n", charVar) is sufficient.
scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input - and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x - which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn't need to be preceded by an ampersand.
I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. :)
%s is for reading a string -- multiple characters delimited by whitespace. %c is for reading a single char.
You declare your yourname and yoursex vars as characters, and then try to read strings into them. The string read will overwrite random other things in the stack frame and misbehave or crash.
You want to declare yourname and yoursex as character arrays, so they can hold strings:
char yourname[32];
char yoursex[32];
then, when reading into them, you want to include a length limit so they don't overflow:
scanf("%31s", yourname);
This is a single character:
char yourname;
But %s indicates that the variable is a string (i.e., an array of characters terminated by a NUL). That's why you need %c. If you really did mean to use a string, then define the variable like
char yourname[32]; /* just pick a big enough size */
Also, you are correct to use the address of the variable with scanf(), but printf() needs the value. So instead of
printf("I see you are %d, you have many years then!", &yourage);
use
printf("I see you are %d, you have many years then!", yourage);
The "big number" is the memory address.
Make sure you read the comments in code!
#include <stdio.h>
int main()
{
char yourname[10];
char yoursex[5]; // boy or girl + null terminator
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &(*yourname)); // & and * cancel each other out,
// thus take a look at the next scanf()
printf("Oh, hello %s! \n\n", yourname); // yourname is now an array
printf("Are you a boy or a girl?: ");
scanf("%s", yoursex);
printf("Nice to know you are a %s! \n\n", yoursex);
printf("How old are you %s? I am ", yourname);
scanf("%d", &yourage); // ok
printf("I see you are %d, you have many years then!", yourage); // here you don't
// need the address of the variable!
return 0;
}
The expression char yourname; only holds space for a single character, so quite likely you end up corrupting the memory space when scanning for yourname. You should allocate a bigger buffer and make sure that you don't overrun its length by setting a maximum number of characters to be read with the scanf function; as described in some of the other answers.
The fact that the following printf print correctly the name doesn't mean that the memory doesn't get corrupted; as C/C++ don't really check the boundary of any strings or arrays used at runtime.
As suggested by others, starting by reading a good book about C and/or C++ wouldn't a bad idea.
int main()
{
//Define Variables
char studentName;
//Print instructions to fill the data in the screen
printf("Please type in the Students name:\n");
scanf("%s", &studentName);
printf("\n\n%s", &studentName);
return 0;
}
Seeing the above code, I am only printing to screen out the first word when I type in a sentence.
I know it is a basic thing, but I am just starting with plain C.
Read scanf(3) documentation. For %s is says
s Matches a sequence of non-white-space characters; the next
pointer must be a pointer to character array that is long
enough to hold the input sequence and the terminating null
byte ('\0'), which is added automatically. The input string
stops at white space or at the maximum field width, whichever
occurs first.
So your code is wrong, because it should have an array for studentName i.e.
char studentName[32];
scanf("%s", studentName);
which is still dangerous because of possible buffer overflow (e.g. if you type a name of 32 or more letters). Using %32s instead of %s might be safer.
Take also the habit of compiling with all warnings enabled and with debugging information (i.e. if using GCC with gcc -Wall -g). Some compilers might have warned you. Learn to use your debugger (such as gdb).
Also, take the habit of ending -not starting- your printf format string with \n (or else call fflush, see fflush(3)).
Learn about undefined behavior. Your program had some! And it misses a #include <stdio.h> directive (as the first non-comment significant line).
BTW, reading existing free software code in C will also teach you many things.
There are three problems with your code:
You are writing a string into a block of memory allocated for a single character; this is undefined behavior
You are printing a string from a block of memory allocated for a single character - also an undefined behavior
You are using scanf to read a string with spaces; %s stops at the first space or end-of-line character.
One way to fix this would be using fgets, like this:
char studentName[100];
//Print instructions to fill the data in the screen
printf("Please type in the Students name:\n");
fgets(studentName, 100, stdin);
printf("\n\n%s", &studentName);
return 0;
Try scanf("%[^\n]", &studentName); instead of scanf("%s", &studentName);
This is happening because %s stops reading the input as soon as a white space is encountered.
To avoid this what you can do is declare an array of the length required for your string.
Then use this command to input the string:-
scanf("%[^\n]s",arr);
This way scanf will continue to read characters unless a '\n' is encountered, in other words you press the enter key on your keyboard. This gives a new line signal and the input stops.
int main()
{
//Define Variables
char studentName[50];
//Print instructions to fill the data in the screen
printf("Please type in the Students name:\n");
scanf("%[^\n]s", &studentName);
printf("\n\n%s", &studentName);
return 0;
}
Alternatively you can also use the gets() and puts() method. This will really ease your work if you are writing a code for a very basic problem.
[EDIT] : As dasblinkenlight has pointed out...I will also not recommend you to use the gets function since it has been deprecated.
int main()
{
//Define Variables
char studentName[50];
//Print instructions to fill the data in the screen
printf("Please type in the Students name:\n");
gets(studentName); printf("\n\n");
puts(studentName);
return 0;
}
make the changes below and try it. I added [80] after the studentName definition, to tell the compiler that studentName is an array of 80 characters (otherwise the compiler would treat it as only one char). Also, the & symbol before studentName is not necessary, because the name of the array implicitly implies a pointer.
int main()
{
//Define Variables
char studentName[80];
//Print instructions to fill the data in the screen
printf("Please type in the Students name:\n");
scanf("%s", studentName);
printf("\n\n%s", studentName);
return 0;
}
Your problem is here
char studentName;
It is a char, not a string.
Try:
Define it as an array of chars like char studenName[SIZE];.
allocating memory dynamically using malloc:
.
char buffer[MAX_SIZE];
scanf("%s", &buffer);
char * studentName = malloc (sizeof(buffer) + 1);
strcpy (studentName , buffer);