I'm trying to convert the digit's 0 to 9 to ASCII using _itoa_s and I find myself running into stack corruption errors using MSVC2012.
I thought that the ASCII table only occupied one byte per character, but from the looks of things, one byte isn't enough.
Where is my thinking wrong?
for (int digit = 0; digit < 10; digit++)
{
char ch_digit;
_itoa_s(digit, &ch_digit, 1, 10);
}
I thought this simple loop should succeed, however it fails. I'm baffled.
_itoa_s() should write out one char AND the terminating NUL char. You want it to write into a buffer with length = 1. So either you get the corruption error because of uninitialized ch_digit, or _itoa_s() is not _s (secure) and shmashes your stack by writing behind that one char.
But why not just calculating the ASCII char for base 10 'by hand' instead of using this non portable, MS specific lumber ?
for (int digit = 0; digit < 10; digit++)
{
char ch_digit = '0' + digit; //'0' == 48 == 0x30
}
itoa_* writes a string, so there's a null-terminator involved.
Related
I am creating a simple encryption program.
I am adding chars to chars to create a new char.
As of now the new 'char' is often a represented by a '?'.
My assumption was that the char variable has a max sum and once it was passed it looped back to 0.
assumed logic:
if char a == 1 && char z == 255
then 256 should == a.
This does not apear to be the case.
This snippet adds a char to a char.
It often prints out something like:
for (int i = 0; i < half; ++i) {
halfM1[i] = halfM1[i] + halfP1[i];
halfM2[i] = halfM2[i] + halfP2[(half + i)];
}
printf("\n%s\n", halfM1 );
printf("%s\n", halfM2);
Returns:
a???
3d??
This snippet removes the added char and the strings go back to normal.
for (int i = 0; i < half; ++i) {
halfM1[i] = halfM1[i] - halfP1[i];
halfM2[i] = halfM2[i] - halfP2[(half + i)];
}
printf("\n%s\n", halfM1 );
printf("%s\n", halfM2);
returns:
messagepart1
messagepart2
The code technically works, but I would like the encryption to be in chars.
If question on why 'half' is everywhere.
The message and key are split in half so the first half and second half of message have separate encryption.
First of all, there is no such thing as "wraparound" for common char. A common char is a signed type in x86, and signed integers do not have wraparound. Instead the overflow leads to undefined behaviour. Additionally, the range of chars can be -128 ... 127, or even something
For cryptographic purposes you'd want to use unsigned chars, or even better, raw octets with uint8_t (<stdint.h>).
Second problem is that you're printing with %s. One of the possible 256 resulting characters is \0. If this gets into the resulting string, it will terminate the string prematurely. Instead of using %s, you should output it with fwrite(halfM1, buffer_size, 1, stdout). Of course the problem is that the output is still some binary garbage. For this purposes many Unix encryption programs will write to file, or have an option to output an ASCII-armoured file. A simple ASCII armouring would be to output as hex instead of binary.
The third is that there is an operation that is much better than addition/subtraction for cryptographic purposes: XOR, or halfM1[i] = halfM1[i] ^ halfP1[i]; - the beauty of which is that it is its own inverse!
This is probably a very simple solution, but for the life of me I can't figure it out. I'm trying to create a char array (so a char*) consisting of numbers from 0 to numPlayers - 1, which I will iterate through to access whose turn it is. So, if numPlayers = 10, I want gameState.players to be [0,1,2,3,4,5,6,7,8,9]. What did I do wrong?
printf("How many players will be playing: ");
scanf(" %d", &numPlayers);
gameState.players = (char*) malloc(numPlayers * sizeof(char));
for (int i = 0; i < numPlayers; ++i) {
strcpy(gameState.players[i],(char) i);
}
First off:
gameState.players = (char*) malloc(numPlayers * sizeof(char));
The explicit cast is ill-advised in C (it can hide certain subtle errors) and the multiplication by sizeof(char) is never needed - it's always one.
But the real problem lies here:
strcpy(gameState.players[i],(char) i);
The str* functions are meant to work with C strings (null terminated character arrays). You do not have a string, rather you have a character value, so it should be more along the lines of:
gameState.players[i] = i;
You also need to keep in mind:
Though you're using char variables, the value being put in is not the textual representation of the digit. To get that, you would need to use i + '0'(a). Characters are generally meant to be used for (mostly) printable stuff, you would be better off using a more-specific data type like int or unsigned short` for non-character data.
This scheme (assuming you want textual representation) is going to break horribly if you ever use more than ten items.
(a) There's a big difference between the "characters" 7 and '7'. The former actually has the value 7 (ASCII BEL, if you're using ASCII), the latter has the value 0x37 (again, assuming ASCII/Unicode).
The numeric characters are the only ones guaranteed to be consecutive so you can convert a numeric value 0..9 to the printable character value simply by adding '0'.
Please be careful when calling functions, you have to make sure that you are
using the correct types. If you use an incorrect one, the compiler will warn you
or print an error, you should read the warnings and errors of the compiler. They
tell you what is wrong.
strcpy is used to copy strings. The signature of the functions is
char *strcpy(char *dest, const char *src);
it expects a pointer to char as destination, and a pointer to char as the
source. Also strcpy expects a valid C-String. A C-Strings is a sequence of
char bytes that ends with the value '\0'. If you don't have a valid
C-String, you cannot use strcpy.
strcpy(gameState.players[i],(char) i)
Is wrong on many levels:
The arguments are not pointers of char
You are not dealing with valid C-strings
Casting won't help you here, you even did the wrong casting.
1 is not the same as '1'. The character '1' is actually the vakue 49
(ASCII code). If you want to get the ASCII representation of a digit, you have
to do: '0' + digit.
You should do:
for (int i = 0; i < numPlayers; ++i) {
gameState.players[i] = '0' + i;
}
Note that this would only work for max. 9 players. If you need more, then you
have to use an array of strings.
Also note that gameState.players does not point to a C-String, because it is
not '\0'-terminated. You cannot calls strings functions on it. If you want
to do that, then you have to change your code like this:
printf("How many players will be playing: ");
fflush(stdout);
scanf(" %d", &numPlayers);
if(numPlayers > 9)
{
// error, do not continue
}
gameState.players = calloc(numPlayers + 1, sizeof *gameState.players);
for (int i = 0; i < numPlayers; ++i) {
gameState.players[i] = '0' + i;
}
int k=5;
char* result = (char *)malloc(100 * sizeof(char));
result[count] = k;
Considering the above code, if I print the contents of result[count] it prints smiley symbols. I tried like below,but of no use.
result[count]=(char)k;
Can anyone help me?
I am percepting that malloc(100*sizeof(char)) will create 100 blocks each of size of character contiguously. Please correct me if I am wrong?
I'll disregard all the fluff about arrays and malloc as it is irrelevant to the problem you describe. You seem to essentially be asking "What will this code print:"
char c = 5;
printf("%c", c);
It will print the symbol 5 in your symbol table, most likely a non-printable character. When encountering non-printable characters, some implementations choose to print non-standard symbols, such as smileys.
If you want to print the number 5, you have to use a character literal:
char c = '5';
printf("%c", c);
or alternatively use poor style with "magic numbers":
char c = 53; // don't write code like this
printf("%c", c);
It is a problem of character representation. Let's begin by the opposite first. A char can be promoted to an int, but if you do you get the representation of the character (its code). Assuming you use ASCII:
char c = '5';
int i = c; // i is now 0x35 or 53 ASCII code for '5'
Your example is the opposite, 5 can be represented by a char, so result[count] = k is defined (even if you should have got a warning for possible truncation), but ASCII char for code 5 is control code ENQ and will be printed as ♣ on a windows system using code page 850 or 437.
A half portable (not specified, but is known to work on all common architectures) way to do the conversion (int) 5 -> (char) '5' is to remember that code '0' to '9' are consecutive (as are 'A' to 'Z' and 'a to 'z'), so I think that what you want is:
result[count] = '0' + k;
First, let's clean up your code a bit:
int k=5;
char* result = malloc(100);
result[count] = k;
sizeof(char) is always equal to 1 under any platform and you don't need to cast the return of malloc() in C.
This said, as long as your count variable contains a value between 0 and 99, you're not doing anything wrong. If you want to print the actual character '5', then you have to assign the ASCII value of that character (53), not just the number 5. Also, remember that if that array of chars has to be interpreted as a string, you need to terminate it with '\0':
int k=5, i;
char* result = malloc(100);
for (i=0; i<98; i++) {
result[i] = ' ';
}
result[98] = 53;
result[99] = '\0';
printf("%s\n", result);
The elements in the array you're trying to allocate are all the size of a char, 1 byte. An int is 4 bytes thus your issues. If you want to store ints in an array you could do:
int result[100];
Followed by storing the value, but honestly from the way your questioned is posed I don't know if that's actually what you're trying to do. Try rephrasing the post with what you're trying to accomplish.
Are you trying to store 5 as a character?
I'm trying to add an int to a char array. My (broken) code is as follows,
string[i] = (char) number;
with i being some int index of the array and number is some integer number. Actually, while typing this out I noticed another problem that would occur if the number is more than one digit, so if you have some answer to that problem as well that would be fantastic!
Given the revised requirement to get digit '0' into string[i] if number == 0, and similarly for values of number between 1 and 9, then you need to add '0' to the number:
assert(number >= 0 && number <= 9);
string[i] = number + '0';
The converse transform is used to convert a digit character back to the corresponding number:
assert(isdigit(c));
int value = c - '0';
If you want to convert a single digit to a number character you can use
string[i] = (char) (number+'0');
Of course you should check if the int value is between 0 and 9. If you have arbitrary numbers and you want to convert them to a string, you should use snprintf, but of course, then you can't squeeze it in a char aynmore, because each char represents a single digit.
If you create the digit representation by doing it manually, you should not forget that a C string requires a \0 byte at the end.
You'll want to use sprintf().
sprintf(string,'%d',number);
I believe.
EDIT: to answer the second part of your question, you're casting an integer to a character, which only holds one digit, as it were. You'd want to put it in a char* or an array of chars.
use asprintf :
char *x;
int size = asprintf(&x, "%d", number);
free(x);
is better because you don't have to allocate memory. is done by asprintf
It is a very trivial question but I don't know why I am not getting the correct output. Here is what I am trying to do:
char sendBuffer[1000];
int count=0:
while(count<10)
{
sendBuffer[0]=count;
sendBuffer[1]='a';
sendBuffer[2]='b';
sendBuffer[3]='c';
sendBuffer[4]='\0';
printf(%s,sendBuffer);
count=count+1;
}
In the output, all buffer is printed correctly except the first index. I want 1,2,3 and so on to be printed at the start of the buffer but it does not work. Please Help
You need to convert that number into a character. A cheap way to do it is:
sendBuffer[0] = '0' + count;
is there anyway to display integers greater than 9
If you need that you'll want to shift to more elaborate schemes. For example, if you want to convert an integer 42 into the string "42" you can say:
#define ENOUGH ((CHAR_BIT * sizeof(int) - 1) / 3 + 2)
char str[ENOUGH];
snprint(str, sizeof str, "%d", 42);
Credit for ENOUGH goes to caf.
printf(%s,sendBuffer); should be printf("%s",sendBuffer);
Change
sendBuffer[0]=count;
to
sendBuffer[0]='0' + count;
i.e. convert the integer 0...9 to the characters '0' ... '9'
Also add a quote i.e. printf("%s",sendBuffer);
Quoted from the question:
In the output, all buffer is printed correctly except the first
index.i want 1,2,3 and so on to be printed at the start of the buffer
but it does not work. Please Help
I think her problem is that she does not get any output for the first line. That is because in the first iteration, count is 0, i.e. sendBuffer[0] is 0 which is the '\0' character. Hence the string is treated as empty.
You are trying to print the ascii characters corresponding to decimal values 0-9 which are non-printable characters. If you need to print decimal 0-9 then initialise count to 48 which is the ascii decimal code for '0' and change the condition in the while block to count < 58; 57 is the ascii deciaml code for '9'. See below:
char sendBuffer[1000];
int count=48:
while(count<58)
{
sendBuffer[0]=count;
sendBuffer[1]='a';
sendBuffer[2]='b';
sendBuffer[3]='c';
sendBuffer[4]='\0';
printf(%s,sendBuffer);
count=count+1;
}
To convert an integer value to a char representation, add the value of the character '0':
sendBuffer[0]=count + '0';
Notice that's the character '0', not the number 0. This is because of how ascii values work. The char with a literal value of 0 is \0, the null terminator. Digit '0' has a literal value of 48, '1' 49 and so on.
This works for the digits 0-9. You can't put a char representation of a number bigger than that in a single char, obviously.
char sendBuffer[1000];
// take fixed stuff outside the loop
sendBuffer[1]='a';
sendBuffer[2]='b';
sendBuffer[3]='c';
sendBuffer[4]='\0';
// it's best not to rely on order of ASCII values
for(char* numbers="0123456789"; *numbers!=0; numbers++){// use for loop for numbers
sendBuffer[0] = *numbers;
printf("%s",sendBuffer);
}