Can someone explain the output of this simple program?
#include <stdio.h>
int main(int argc, char *argv[])
{
char charArray[1024] = "";
char charArrayAgain[1024] = "";
int number;
number = 2;
sprintf(charArray, "%d", number);
printf("charArray : %s\n", charArray);
snprintf(charArrayAgain, 1, "%d", number);
printf("charArrayAgain : %s\n", charArrayAgain);
return 0;
}
The output is:
./a.out
charArray : 2
charArrayAgain : // Why isn't there a 2 here?
Because snprintf needs space for the \0 terminator of the string. So if you tell it the buffer is 1 byte long, then there's no space for the '2'.
Try with snprintf(charArrayAgain, 2, "%d", number);
snprintf(charArrayAgain, 1, "%d", number);
// ^
You're specifying your maximum buffer size to be one byte. However, to store a single digit in a string, you must have two bytes (one for the digit, and one for the null terminator.)
You've told snprintf to only print a single character into the array, which is not enough to hold the string-converted number (that's one character) and the string terminator \0, which is a second character, so snprintf is not able to store the string into the buffer you've given it.
The second argument to snprintf is the maximum number of bytes to be written to the array (charArrayAgain). It includes the terminating '\0', so with size of 1 it's not going to write an empty string.
Check the return value from the snprintf() it will probably be 2.
Related
I'm trying to use sprintf() to put a string "inside itself", so I can change it to have an integer prefix. I was testing this on a character array of length 12 with "Hello World" inside it already.
The basic premise is that I want a prefix that denotes the amount of words within a string. So I copy 11 characters into a character array of length 12.
Then I try to put the integer followed by the string itself by using "%i%s" in the function. To get past the integer (I don't just use myStr as the argument for %s), I make sure to use myStr + snprintf(NULL, 0, "%i", wordCount), which should be myStr + characters taken up by the integer.
The problem is that I'm having is that it eats the 'H' when I do this and prints "2ello World" instead of having the '2' right beside the "Hello World"
So far I've tried different options for getting "past the integer" in the string when I try to copy it inside itself, but nothing really seems to be the right case, as it either comes out as an empty string or just the integer prefix itself '222222222222' copied throughout the entire array.
int main() {
char myStr[12];
strcpy(myStr, "Hello World");//11 Characters in length
int wordCount = 2;
//Put the integer wordCount followed by the string myStr (past whatever amount of characters the integer would take up) inside of myStr
sprintf(myStr, "%i%s", wordCount, myStr + snprintf(NULL, 0, "%i", wordCount));
printf("\nChanged myStr '%s'\n", myStr);//Prints '2ello World'
return 0;
}
First, to insert a one-digit prefix into a string “Hello World”, you need a buffer of 13 characters—one for the prefix, eleven for the characters in “Hello World”, and one for the terminating null character.
Second, you should not pass a buffer to snprintf as both the output buffer and an input string. Its behavior is not defined by the C standard when objects passed to it overlap.
Below is a program that shows you how to insert a prefix by moving the string with memmove. This is largely tutorial, as it is not generally a good way to manipulate strings. For short strings, where space is not an issue, most programmers would simply print the desired string into a temporary buffer, avoiding overlap issues.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
/* Insert a decimal numeral for Prefix into the beginning of String.
Length specifies the total number of bytes available at String.
*/
static void InsertPrefix(char *String, size_t Length, int Prefix)
{
// Find out how many characters the numeral needs.
int CharactersNeeded = snprintf(NULL, 0, "%i", Prefix);
// Find the current string length.
size_t Current = strlen(String);
/* Test whether there is enough space for the prefix, the current string,
and the terminating null character.
*/
if (Length < CharactersNeeded + Current + 1)
{
fprintf(stderr,
"Error, not enough space in string to insert prefix.\n");
exit(EXIT_FAILURE);
}
// Move the string to make room for the prefix.
memmove(String + CharactersNeeded, String, Current + 1);
/* Remember the first character, because snprintf will overwrite it with a
null character.
*/
char Temporary = String[0];
// Write the prefix, including a terminating null character.
snprintf(String, CharactersNeeded + 1, "%i", Prefix);
// Restore the first character of the original string.
String[CharactersNeeded] = Temporary;
}
int main(void)
{
char MyString[13] = "Hello World";
InsertPrefix(MyString, sizeof MyString, 2);
printf("Result = \"%s\".\n", MyString);
}
The best way to deal with this is to create another buffer to output to, and then if you really need to copy back to the source string then copy it back once the new copy is created.
There are other ways to "optimise" this if you really needed to, like putting your source string into the middle of the buffer so you can append and change the string pointer for the source (not recommended, unless you are running on an embedded target with limited RAM and the buffer is huge). Remember code is for people to read so best to keep it clean and easy to read.
#define MAX_BUFFER_SIZE 128
int main() {
char srcString[MAX_BUFFER_SIZE];
char destString[MAX_BUFFER_SIZE];
strncpy(srcString, "Hello World", MAX_BUFFER_SIZE);
int wordCount = 2;
snprintf(destString, MAX_BUFFER_SIZE, "%i%s", wordCount, srcString);
printf("Changed string '%s'\n", destString);
// Or if you really want the string put back into srcString then:
strncpy(srcString, destString, MAX_BUFFER_SIZE);
printf("Changed string in source '%s'\n", srcString);
return 0;
}
Notes:
To be safer protecting overflows in memory you should use strncpy and snprintf.
I am trying to make function that compares all the letters from alphabet to string I insert, and prints letters I didn't use. But when I print those letters it goes over and gives me random symbols at end. Here is link to function, how I call the function and result: http://imgur.com/WJRZvqD,U6Z861j,PXCQa4V#0
Here is code: (http://pastebin.com/fCyzFVAF)
void getAvailableLetters(char lettersGuessed[], char availableLetters[])
{
char alphabet[]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
int LG,LG2,LA=0;
for (LG=0;LG<=strlen(alphabet)-1;LG++)
{
for(LG2=0;LG2<=strlen(lettersGuessed)-1;LG2++)
{
if (alphabet[LG]==lettersGuessed[LG2])
{
break;
}
else if(alphabet[LG]!=lettersGuessed[LG2] &&LG2==strlen(lettersGuessed)-1)
{
availableLetters[LA]=alphabet[LG];
LA++;
}
}
}
}
Here is program to call the function:
#include <stdio.h>
#include <string.h>
#include "hangman.c"
int main()
{
int i = 0;
char result[30];
char text[30];
scanf("%s", text);
while(i != strlen(text))
{
i++;
}
getAvailableLetters(text, result);
printf("%s\n", result);
printf ("%d", i);
printf ("\n");
}
Here is result when I typed in abcd: efghijklmnopqrstuvwxyzUw▒ˉ
If you want to print result as a string, you need to include a terminating null at the end of it (that's how printf knows when to stop).
for %s printf stops printing when it reaches a null character '\0', because %s expects the string to be null terminated, but result not null terminated and that's why you get random symbols at the end
just add availableLetters[LA] = '\0' at the last line in the function getAvailableLetters
http://pastebin.com/fCyzFVAF
Make sure your string is NULL-terminated (e.g. has a '\0' character at the end). And that also implies ensuring the buffer that holds the string is large enough to contain the null terminator.
Sometimes one thinks they've got a null terminated string but the string has overflowed the boundary in memory and truncated away the null-terminator. That's a reason you always want to use the form of functions (not applicable in this case) that read data, like, for example, sprintf() which should be calling snprintf() instead, and any other functions that can write into a buffer to be the form that let's you explicitly limit the length, so you don't get seriously hacked with a virus or exploit.
char alphabet[]={'a','b','c', ... ,'x','y','z'}; is not a string. It is simply an "array 26 of char".
In C, "A string is a contiguous sequence of characters terminated by and including the first null character. ...". C11 §7.1.1 1
strlen(alphabet) expects a string. Since code did not provide a string, the result is undefined.
To fix, insure alphabet is a string.
char alphabet[]={'a','b','c', ... ,'x','y','z', 0};
// or
char alphabet[]={"abc...xyz"}; // compiler appends a \0
Now alphabet is "array 27 of char" and also a string.
2nd issue: for(LG2=0;LG2<=strlen(lettersGuessed)-1;LG2++) has 2 problems.
1) Each time through the loop, code recalculates the length of the string. Better to calculate the string length once since the string length does not change within the loop.
size_t len = strlen(lettersGuessed);
for (LG2 = 0; LG2 <= len - 1; LG2++)
2) strlen() returns the type size_t. This is some unsigned integer type. Should lettersGuessed have a length of 0 (it might have been ""), the string length - 1 is not -1, but some very large number as unsigned arithmetic "wraps around" and the loop may never stop. A simple solution follows. This solution would only fail is the length of the string exceeded INT_MAX.
int len = (int) strlen(lettersGuessed);
for (LG2 = 0; LG2 <= len - 1; LG2++)
A solution without this limitation would use size_t throughout.
size_t LG2;
size_t len = strlen(lettersGuessed);
for (LG2 = 0; LG2 < len; LG2++)
I am currently learning C, and so I wanted to make a program that asks the user to input a string and to output the number of characters that were entered, the code compiles fine, when I enter just 1 character it does fine, but when I enter 2 or more characters, no matter what number of character I enter, it will always say there is just one character and crashes after that. This is my code and I can't figure out what is wrong.
int main(void)
{
int siz;
char i[] = "";
printf("Enter a string.\n");
scanf("%s", i);
siz = sizeof(i)/sizeof(char);
printf("%d", siz);
getch();
return 0;
}
I am currently learning to program, so if there is a way to do it using the same scanf() function I will appreciate that since I haven't learned how to use any other function and probably won't understand how it works.
Please, FORGET that scanf exists. The problem you are running into, whilst caused mostly by your understandable inexperience, will continue to BITE you even when you have experience - until you stop.
Here is why:
scanf will read the input, and put the result in the char buffer you provided. However, it will make no check to make sure there is enough space. If it needs more space than you provided, it will overwrite other memory locations - often with disastrous consequences.
A safer method uses fgets - this is a function that does broadly the same thing as scanf, but it will only read in as many characters as you created space for (or: as you say you created space for).
Other observation: sizeof can only evaluate the size known at compile time : the number of bytes taken by a primitive type (int, double, etc) or size of a fixed array (like int i[100];). It cannot be used to determine the size during the program (if the "size" is a thing that changes).
Your program would look like this:
#include <stdio.h>
#include <string.h>
#define BUFLEN 100 // your buffer length
int main(void) // <<< for correctness, include 'void'
{
int siz;
char i[BUFLEN]; // <<< now you have space for a 99 character string plus the '\0'
printf("Enter a string.\n");
fgets(i, BUFLEN, stdin); // read the input, copy the first BUFLEN characters to i
siz = sizeof(i)/sizeof(char); // it turns out that this will give you the answer BUFLEN
// probably not what you wanted. 'sizeof' gives size of array in
// this case, not size of string
// also not
siz = strlen(i) - 1; // strlen is a function that is declared in string.h
// it produces the string length
// subtract 1 if you don't want to count \n
printf("The string length is %d\n", siz); // don't just print the number, say what it is
// and end with a newline: \n
printf("hit <return> to exit program\n"); // tell user what to do next!
getc(stdin);
return 0;
}
I hope this helps.
update you asked the reasonable follow-up question: "how do I know the string was too long".
See this code snippet for inspiration:
#include <stdio.h>
#include <string.h>
#define N 50
int main(void) {
char a[N];
char *b;
printf("enter a string:\n");
b = fgets(a, N, stdin);
if(b == NULL) {
printf("an error occurred reading input!\n"); // can't think how this would happen...
return 0;
}
if (strlen(a) == N-1 && a[N-2] != '\n') { // used all space, didn't get to end of line
printf("string is too long!\n");
}
else {
printf("The string is %s which is %d characters long\n", a, strlen(a)-1); // all went according to plan
}
}
Remember that when you have space for N characters, the last character (at location N-1) must be a '\0' and since fgets includes the '\n' the largest string you can input is really N-2 characters long.
This line:
char i[] = "";
is equivalent to:
char i[1] = {'\0'};
The array i has only one element, the program crashes because of buffer overflow.
I suggest you using fgets() to replace scanf() like this:
#include <stdio.h>
#define MAX_LEN 1024
int main(void)
{
char line[MAX_LEN];
if (fgets(line, sizeof(line), stdin) != NULL)
printf("%zu\n", strlen(line) - 1);
return 0;
}
The length is decremented by 1 because fgets() would store the new line character at the end.
The problem is here:
char i[] = "";
You are essentially creating a char array with a size of 1 due to setting it equal to "";
Instead, use a buffer with a larger size:
char i[128]; /* You can also malloc space if you desire. */
scanf("%s", i);
See the link below to a similar question if you want to include spaces in your input string. There is also some good input there regarding scanf alternatives.
How do you allow spaces to be entered using scanf?
That's because char i[] = ""; is actually an one element array.
Strings in C are stored as the text which ends with \0 (char of value 0). You should use bigger buffer as others said, for example:
char i[100];
scanf("%s", i);
Then, when calculating length of this string you need to search for the \0 char.
int length = 0;
while (i[length] != '\0')
{
length++;
}
After running this code length contains length of the specified input.
You need to allocate space where it will put the input data. In your program, you can allocate space like:
char i[] = " ";
Which will be ok. But, using malloc is better. Check out the man pages.
Have a problem accessing some elements of a structure in a right way after typecasting.
Here my code:
void get_description(struct shmstruct *ptr/*, int number*/) {
char buff[MESGSIZE];
struct shmData *dparse;
snprintf(buff, MESGSIZE, "%s", &ptr->msgdata[0]);
dparse = (struct shmData *) buff;
printf("Number: %s", dparse->number);
printf("Description: %s", dparse->description);
}
Problem now is, that I get the number, like 123, but also the description in the first
line -> printf("Number: %s", dparse->number);
like: 123 description
How can I get only the number?
(P.S.
struct shmData{
char number[4];
char description[1020];
};
)
It sounds like the number is not nul-terminated. You have the number field and then immediately following you have the description field. printf() assumes that you are giving it a nul-terminated string, and it will keep going until it hits the terminating nul character. It looks like in your case, there isn't a nul terminating the number field, so printf() just keeps going and gets the description also.
Can all four characters from number be used for digits?
If the numbers are only 3-digits or fewer, then you can put a terminating nul right into the number buffer. If you might need to read a 4-digit number from there, you need to copy out the digits into a temp buffer of at least 5 chars, and then nul-terminate.
// max 3 digits:
dparse->number[3] = '\0'; // ensure nul-termination of number
// max 4 digits:
#define DPARSE_NUMBER 4 /* size of number field in dparse */
char temp[DPARSE_NUMBER + 1];
strncpy(temp, dparse->number, sizeof(temp));
temp[sizeof(temp) - 1] = '\0'; // strncpy() does not guarantee to nul-terminate, so we must do this
strncpy() doesn't guarantee to nul-terminate, which is unfortunate. So we should always put in a terminating nul character in the last position so that no matter what, it is nul-terminated. Note that sometimes strncpy() will put a nul for us, but it does no harm to make sure one is there.
I have a function that will parse some data coming in. My problem is that after using strncpy I get some garbage when I try to print it. I try using malloc to make the char array the exact size.
Code:
void parse_data(char *unparsed_data)
{
char *temp_str;
char *pos;
char *pos2;
char *key;
char *data;
const char newline = '\n';
int timestamp = 0;
temp_str = (char*)malloc(strlen(unparsed_data));
g_print("\nThe original string is: \n%s\n",unparsed_data);
//Ignore the first two lines
pos = strchr(unparsed_data, newline);
strcpy(temp_str, pos+1);
pos = strchr(temp_str, newline);
strcpy(temp_str, pos+1);
//Split the line in two; The key name and the value
pos = strchr(temp_str, ':'); // ':' divides the name from the value
pos2 = strchr(temp_str, '\n'); //end of the line
key = (char*)malloc((size_t)(pos-temp_str)-1); //allocate enough memory
data = (char*)malloc((size_t)(pos2-pos)-1);
strncpy(key, temp_str, (size_t)(pos-temp_str));
strncpy(data, pos + 2, (size_t)(pos2-pos));
timestamp = atoi(data);
g_print("size of the variable \"key\" = %d or %d\n", (size_t)(pos-temp_str), strlen(key));
g_print("size of the variable \"data\" = %d or %d\n", (size_t)(pos2-pos), strlen(data));
g_print("The key name is %s\n",key);
g_print("The value is %s\n",data);
g_print("End of Parser\n");
}
Output:
The original string is:
NEW_DATAa_PACKET
Local Data Set 16-byte Universal Key
Time Stamp (microsec): 1319639501097446
Frame Number: 0
Version: 3
Angle (deg): 10.228428
size of the variable "key" = 21 or 22
size of the variable "data" = 18 or 21
The key name is Time Stamp (microsec)
The value is 1319639501097446
F32
End of Parser
Run it again:
The original string is:
NEW_DATAa_PACKET
Local Data Set 16-byte Universal Key
Time Stamp (microsec): 1319639501097446
Frame Number: 0
Version: 3
Angle (deg): 10.228428
size of the variable "key" = 21 or 25
size of the variable "data" = 18 or 18
The key name is Time Stamp (microsec)ipe
The value is 1319639501097446
F
End of Parser
Your results are because strncpy does not put a null character at the end of the string.
Your strncpy(data, pos + 2, (size_t)(pos2-pos)); doesn't add a terminating \0 character at the end of the string. Therefore when you try to print it later, printf() prints your whole data string and whatever is in the memory right after it, until it reaches zero - that's the garbate you're getting. You need to explicitly append zero at the end of your data. It's also needed for atoi().
Edit:
You need to allocate one more byte for your data, and write a terminating character there. data[len_of_data] = '\0'. Only after that it becomes a valid C string and you can use it for atoi() and printf().
You need to malloc() +1 byte for a string, so it can append the zero when you do strcpy(), but the strncpy will not append zero as well you need a extra byte for it.
One problem: What if there's no newline?
Undefined behaviour:
pos = strchr(temp_str, newline);
strcpy(temp_str, pos+1);
The source and destination of strcpy must not overlap.
You have to remember in allocating space for a string to add one byte for the terminating '\0' character. You have to be careful with strncpy, especially if you are used to using strcpy, strcat, or sprintf. Those three functions terminate the string with '\0'. strncpy copies a number of bytes that you specify, and makes no assumption of terminating the string.
You assume that responsibility by making sure you place a '\0' at the end of the character buffer to which you copied. That means you have to know the starting the position and the length of the copy and put a '\0' one byte past the sum of the starting position and length.
I chose to solve a sample problem slightly differently, but it still involves knowing the length of what I copied.
In this case, I use strncpy to take the first 9 characters from pcszTestStr1 and
copy them to szTestBuf. Then, I use strcpy -- which terminates the string with a zero --
to append the new part of the sentence.
#include <stdio.h>
#include <string.h>
int n;
int argv_2;
char szTestBuf[100] = {0};
char * pcszTestStr1 =
"This is a very long, long string to be used in a C example, OK?";
int main(int argc, char *argv[])
{
int rc = 0;
printf("The following sentence is too long.\n%s\n", pcszTestStr1);
strncpy(szTestBuf, pcszTestStr1, 9);
strcpy(szTestBuf + 9, " much shorter sentence.");
printf("%s\n", szTestBuf);
return rc;
}
Here's the output of running test.c compiled gcc -o test test.c.
cnorton#hiawatha:~/scratch$ ./test
The following sentence is too long.
This is a very long, long string to be used in a C example, OK?
This is a much shorter sentence.
cnorton#hiawatha:~/scratch$