I'm pretty new to C, and I'm trying to write a function that takes a user input RAM size in B, kB, mB, or gB, and determines the address length. My test program is as follows:
int bitLength(char input[6]) {
char nums[4];
char letters[2];
for(int i = 0; i < (strlen(input)-1); i++){
if(isdigit(input[i])){
memmove(&nums[i], &input[i], 1);
} else {
//memmove(&letters[i], &input[i], 1);
}
}
int numsInt = atoi(nums);
int numExponent = log10(numsInt)/log10(2);
printf("%s\n", nums);
printf("%s\n", letters);
printf("%d", numExponent);
return numExponent;
}
This works correctly as it is, but only because I have that one line commented out. When I try to alter the 'letters' character array with that line, it changes the 'nums' character array to '5m2'
My string input is '512mB'
I need the letters to be able to tell if the user input is in B, kB, mB, or gB.
I am confused as to why the commented out line alters the 'nums' array.
Thank you.
In your input 512mB, "mB" is not digit and is supposed to handled in commented code. When handling those characters, i is 3 and 4. But because length of letters is only 2, when you execute memmove(&letters[i], &input[i], 1);, letters[i] access out of bounds of array so it does undefined behaviour - in this case, writing to memory of nums array.
To fix it, you have to keep unique index for letters. Or better, for both nums and letters since i is index of input.
There are several problems in your code. #MarkSolus have already pointed out that you access letters out-of-bounds because you are using i as index and i can be more than 1 when you do the memmove.
In this answer I'll address some of the other poroblems.
string size and termination
Strings in C needs a zero-termination. Therefore arrays must be 1 larger than the string you expect to store in the array. So
char nums[4]; // Can only hold a 3 char string
char letters[2]; // Can only hold a 1 char string
Most likely you want to increase both arrays by 1.
Further, your code never adds the zero-termination. So your strings are invalid.
You need code like:
nums[some_index] = '\0'; // Add zero-termination
Alternatively you can start by initializing the whole array to zero. Like:
char nums[5] = {0};
char letters[3] = {0};
Missing bounds checks
Your loop is a for-loop using strlen as stop-condition. Now what would happen if I gave the input "123456789BBBBBBBB" ? Well, the loop would go on and i would increment to values ..., 5, 6, 7, ... Then you would index the arrays with a value bigger than the array size, i.e. out-of-bounds access (which is real bad).
You need to make sure you never access the array out-of-bounds.
No format check
Now what if I gave an input without any digits, e.g. "HelloWorld" ? In this case nothin would be written to nums so it will be uninitialized when used in atoi(nums). Again - real bad.
Further, there should be a check to make sure that the non-digit input is one of B, kB, mB, or gB.
Performance
This is not that important but... using memmove for copy of a single character is slow. Just assign directly.
memmove(&nums[i], &input[i], 1); ---> nums[i] = input[i];
How to fix
There are many, many different ways to fix the code. Below is a simple solution. It's not the best way but it's done like this to keep the code simple:
#define DIGIT_LEN 4
#define FORMAT_LEN 2
int bitLength(char *input)
{
char nums[DIGIT_LEN + 1] = {0}; // Max allowed number is 9999
char letters[FORMAT_LEN + 1] = {0}; // Allow at max two non-digit chars
if (input == NULL) exit(1); // error - illegal input
if (!isdigit(input[0])) exit(1); // error - input must start with a digit
// parse digits (at max 4 digits)
int i = 0;
while(i < DIGITS && isdigit(input[i]))
{
nums[i] = input[i];
++i;
}
// parse memory format, i.e. rest of strin must be of of B, kB, mB, gB
if ((strcmp(&input[i], "B") != 0) &&
(strcmp(&input[i], "kB") != 0) &&
(strcmp(&input[i], "mB") != 0) &&
(strcmp(&input[i], "gB") != 0))
{
// error - illegal input
exit(1);
}
strcpy(letters, &input[i]);
// Now nums and letter are ready for further processing
...
...
}
}
I'm having trouble with the following: I want to take a large number (cca. 15-digit) and turn it into individual digits and store them in an array. I will need these digits further on in the code. The problem is, if I declare the array outside the while loop, it stores the first four values, but then I get a segmentation fault. If I declare it within the loop, the code works as I want it to, but then I don't know how to move the array out of that loop, so that I could use it further. How can I solve this? This is what I've compiled:
unsigned long long card_num;
printf("Enter your credit card number:\n");
scanf("%llu", &card_num);
int count = 0;
while (card_num != 0)
{
int digits[count]; //declaring array into which digits will be stored
digits[count] = card_num%10; // get last digit, store it in array
card_num = card_num/10; //remove last digit
printf("Digit[%i] = %i\n", count, digits[count]);
printf("Number of digits: %i\n", count);
count ++;
}
In your code, for the very first iteration
int digits[count];
count is 0, which violates the constraints mentioned in spec. As per C11, chapter 6.7.5.2,
In addition to optional type qualifiers and the keyword static, the [ and ] may delimit an expression or *. If they delimit an expression (which specifies the size of an array), the expression shall have an integer type. If the expression is a constant expression, it shall have a value greater than zero. [....]
and
If the size is an expression that is not an integer constant expression: if it occurs in a declaration at function prototype scope, it is treated as if it were replaced by *; otherwise, each time it is evaluated it shall have a value greater than zero
So, this is not a valid code.
As already mentioned, what you are doing is plain wrong.
There is several ways to solve this issue. The easiest would be to allocate an array at the beginning of your program with enough space for all your usecases with something like :
#define MAX_DIGIT 50
int digits[MAX_DIGIT];
Then you just have to check you are not going over this array by checking that count < MAX_DIGIT.
Another way would be using dynamic memory allocation using an int pointer int *digits and malloc (I let you google that) once you know the size of the array you'll need. You'll have to change a bit your code to know the number of digits before starting to get the digits as you need to allocate the memory before starting to store digits.
You could use realloc to keep a code similar to what you already have, but I wouldn't advise it as it is not efficient to realloc memory for each value that you add.
Your logic is fine.
What went wrong is that you tried to increase the length of a fixed-length array while iterating which is not allowed.
You should never change the length of a fixed-length array anywhere in the program.
However if you want to change the length of an array during runtime you must use the malloc and realloc functions to do so.
Check out this example:
//declarations
int *storeNum;
storeNum = (int *)malloc(1 * sizeof(int));
//logic
while(num != 0) {
if(i > 0)
storeNum = realloc(storeNum, (i+1) * sizeof(int));
storeNum[i] = num % 10;
num = num/10;
++i;
}
Here first I declared the array size initially as one and later incremented it using realloc function.
You also have the array size stored in i which you can use later in your code in loops.
But keep in mind that the digits will be stored in your array in reverse order.
Also, you shouldn't declare an array within a loop.
Since you have "declared" the array, each time the compiler enters the loop while iterating it will consider the array-declaration as a new array. That is all.
Can someone explain to me how the calculation works?
what I don't understand is:
the getch(); function, what does that function does?
2.
Can someone explain to me how the int decimal_binary(int n) operates mathematically?
#include<stdio.h>
int decimal_binary (int n);
void main()
{
int n;
printf("Enter decimal number: ");
scanf("%d", &n);
printf("\n%d", decimal_binary(n));
getch();
}
int decimal_binary(int n)
{
int rem, i = 1, binary = 0;
while(n!=0)
{
rem = n % 2;
n = n/2;
binary = binary + rem*i;
i = i*10;
}
return binary;
}
if for example the n = 10
and this is how i calculate it
I'm not going to explain the code in the question, because I fundamentally (and rather vehemently) disagree with its implementation.
When we say something like "convert a number to base 2", it's useful to understand that we are not really changing the number. All we're doing is changing the representation. An int variable in a computer program is just a number (although deep down inside it's already in binary). The base matters when we print the number out as a string of digit characters, and also when we read it from as a string of digit characters. So any sensible "convert to base 2" function should have as its output a string, not an int.
Now, when you want to convert a number to base 2, and in fact when you want to convert to base b, for any base "b", the basic idea is to repeatedly divide by b.
For example, if we wanted to determine the base-10 digits of a number, it's easy. Consider the number 12345. If we divide it by 10, we get 1234, with a remainder of 5. That remainder 5 is precisely the last digit of the number 12345. And the remaining digits are 1234. And then we can repeat the procedure, dividing 1234 by 10 to get 123 remainder 4, etc.
Before we go any further, I want you to study this base-10 example carefully. Make sure you understand that when we split 12345 up into 1234 and 5 by dividing it by 10, we did not just look at it with our eyes and pick off the last digit. The mathematical operation of "divide by 10, with remainder" really did do the splitting up for us, perfectly.
So if we want to determine the digits of a number using a base other than 10, all we have to do is repeatedly divide by that other base. Suppose we're trying to come up with the binary representation of eleven. If we divide eleven by 2, we get five, with a remainder of 1. So the last bit is going to be 1.
Next we have to work on five. If we divide five by 2, we get two, with a remainder of 1. So the next-to-last bit is going to be 1.
Next we have to work on two. If we divide two by 2, we get one, with a remainder of 0. So the next bit is going to be 0.
Next we have to work on one. If we divide one by 2, we get zero, with a remainder of 1. So the next bit is going to be 1.
And now we have nothing left to work with -- the last division has resulted in 0. The binary bits we've picked off were, in order, 1, 1, 0, and 1. But we picked off the last bit first. So rearranging into conventional left-to-right order, we have 1011, which is the correct binary representation of the number eleven.
So with the theory under our belt, let's look at some actual C code to do this. It's perfectly straightforward, except for one complication. Since the algorithm we're using always gives us the rightmost bit of the result first, we're going to have to do something special in order to end up with the bits in conventional left-to-right order in the final result.
I'm going to write the new code as function, sort of like your decimal_binary. This function will accept an integer, and return the binary representation of that integer as a string. Because strings are represented as arrays of characters in C, and because memory allocation for arrays can be an issue, I'm going to also have the function accept an empty array (passed by the caller) to build the return string in. And I'm also going to have the function accept a second integer giving the size of the array. That's important so that the function can make sure not to overflow the array.
If it's not clear from the explanation so far, here's what a call to the new function is going to look like:
#include <stdio.h>
char *integer_binary(int n, char *str, int sz);
int main()
{
int n;
char result[40];
printf("Enter decimal number: ");
scanf("%d", &n);
char *str = integer_binary(n, result, 40);
printf("%s\n", str);
}
As I said, the new function, integer_binary, is going to create its result as a string, so we have to declare an array, result, to hold that string. We're declaring it as size 40, which should be plenty to hold any 32-bit integer, with some left over.
The new function returns a string, so we're printing its return value using %s.
And here's the implementation of the integer_binary function. It's going to look a little scary at first, but bear with me. At its core, it's using the same algorithm as the original decimal_binary function in the question did, repeatedly dividing by 2 to pick off the bits of the binary number being generated. The differences have to do with constructing the result in a string instead of an int. (Also, it's not taking care of quite everything yet; we'll get to one or two more improvements later.)
char *integer_binary(int n, char *binary, int sz)
{
int rem;
int j = sz - 2;
do {
if(j < 0) return NULL;
rem = n % 2;
n = n / 2;
binary[j] = '0' + rem;
j--;
} while(n != 0);
binary[sz-1] = '\0';
return &binary[j+1];
}
You can try that, and it will probably work for you right out of the box, but let's explain the possibly-confusing parts.
The new variable j keeps track of where in the array result we're going to place the next bit value we compute. And since the algorithm generates bits in right-to-left order, we're going to move j backwards through the array, so that we stuff new bits in starting at the end, and move to the left. That way, when we take the final string and print it out, we'll get the bits in the correct, left-to-right order.
But why does j start out as sz - 2? Partly because arrays in C are 0-based, partly to leave room for the null character '\0' that terminates arrays in C. Here's a picture that should make things clearer. This will be the situation after we've completely converted the number eleven:
0 1 2 31 32 33 34 35 36 37 38 39
+---+---+---+-- ~ --+---+---+---+---+---+---+---+---+---+
result: | | | | ... | | | | | 1 | 0 | 1 | 1 |\0 |
+---+---+---+-- ~ --+---+---+---+---+---+---+---+---+---+
^ ^ ^ ^
| | | |
binary final return initial
j value j
The result array in the caller is declared as char result[40];, so it has 40 elements, from 0 to 39. And sz is passed in as 40. But if we want j to start out "at the right edge" of the array, we can't initialize j to sz, because the leftmost element is 39, not 40. And we can't initialize j as sz - 1, either, because we have to leave room for the terminating '\0'. That's why we initialize j to sz - 2, or 38.
The next possibly-confusing aspect of the integer_binary function is the line
binary[j] = '0' + rem;
Here, rem is either 0 or 1, the next bit of our binary conversion we've converted. But since we're creating a string representation of the binary number, we want to fill the binary result in with one of the characters '0' or '1'. But characters in C are represented by tiny integers, and you can do arithmetic on them. The constant '0' is the value of the character 0 in the machine's character set (typically 48 in ASCII). And the bottom line is that '0' + 1 turns into the character '1'. So '0' + rem turns into '0' if rem is 0, or '1' if rem is 1.
Next to talk about is the loop I used. The original decimal_binary function used while(n != 0) {...}, but I'm using do { ... } while(n != 0). What's the difference? It's precisely that the do/while loop always runs once, even if the controlling expression is false. And that's what we want here, so that the number 0 will be converted to the string "0", not the empty string "". (That wasn't an issue for integer_binary, because it returned the integer 0 in that case, but that was a side effect of its otherwise-poor choice of int as its return value.)
Next we have the line
binary[sz-1] = '\0';
We've touched on this already: it simply fills in the necessary null character which terminates the string.
Finally, there's the last line,
return &binary[j+1];
What's going on there? The integer_binary function is supposed to return a string, or in this case, a pointer to the first character of a null-terminated array of characters. Here we're returning a pointer (generated by the & operator) to the element binary[j+1] in the result array. We have to add one to j because we always subtract 1 from it in the loop, so it always indicates the next cell in the array where we'd store the next character. But we exited the loop because there was no next character to generate, so the last character we did generate was at j's previous value, which is j+1.
(This integer_binary function is therefore mildly unusual in one respect. The caller passes in an empty array, and the function builds its result string in the empty array, but the pointer it returns, which points to the constructed string, does not usually point to the beginning of the passed-in array. It will work fine as long as the caller uses the returned pointer, as expected. But it's unusual, and the caller would get confused if accidentally using its own original result array as if it would contain the result.)
One more thing: that line if(j < 0) return NULL; at the top of the loop is a double check that the caller gave us a big enough array for the result we're generating. If we run out of room for the digits we're generating, we can't generate a correct result, so we return a null pointer instead. (That's likely to cause problems in the caller unless explicitly checked for, but that's a story for another day.)
So integer_binary as discussed so far will work, although I'd like to make three improvements to address some remaining deficiencies:
The decimal_binary function as shown won't handle negative numbers correctly.
The way the decimal_binary function uses the j variable is a bit clumsy. (Evidence of the clumsiness is the fact that I had to expend so many words explaining the j = sz-2 and return &binary[j+1] parts.)
The decimal_binary functions as shown only handles, obviously, binary, but what I really want (although you didn't ask for it) is a function that can convert to any base.
So here's an improved version. Based on the integer_binary function we've already seen, there are just a few small steps to achieve the desired improvements. I'm calling the new function integer_base, because it converts to any base (well, any base up to 10, anyway). Here it is:
char *integer_base(int n, int base, char *result, int sz)
{
int rem;
int j = sz - 1;
int negflag = 0;
if(n < 0) {
n = -n;
negflag = 1;
}
result[j] = '\0';
do {
j--;
if(j < 0) return NULL;
rem = n % base;
n = n / base;
result[j] = '0' + rem;
} while(n != 0);
if(negflag) {
j--;
result[j] = '-';
}
return &result[j];
}
As mentioned, this is just like integer_binary, except:
I've changed the way j is used. Before, it was always the index of the next element of the result array we were about to fill in. Now, it's always one to the right of the next element we're going to fill in. This is a less obvious choice, but it ends up being more convenient. Now, we initialize j to sz-1, not sz-2. Now, we do the decrement j-- before we fill in the next character of the result, not after. And now, we can return &binary[j], without having to remember to subtract 1 at that spot.
I've moved the insertion of the terminating null character '\0' up to the top. Since we're building the whole string right-to-left, it makes sense to put the terminator in first.
I've handled negative numbers, in a kind of brute-force but expedient way. If we receive a negative number, we turn it into a positive number (n = -n) and use our regular algorithm on it, but we set a flag negflag to remind us that we've done so and, when we're all done, we tack a '-' character onto the beginning of the string.
Finally, and this is the biggie, the new function works in any base. It can create representations in base 2, or base 3, or base 5, or base 7, or any base up to 10. And what's really neat is how few modifications were required in order to achieve this. In fact, there were just two: In two places where I had been dividing by 2, now I'm dividing by base. That's it! This is the realization of something I said back at the very beginning of this too-long answer: "The basic idea is to repeatedly divide by b."
(Actually, I lied: There was a fourth change, in that I renamed the result parameter from "binary" to "result".)
Although you might be thinking that this integer_base function looks pretty good, I have to admit that it still has at least three problems:
It won't work for bases greater than 10.
It can occasionally overflow its result buffer.
It has an obscure problem when trying to convert the largest negative number.
The reason it only works for bases up to 10 is the line
result[j] = '0' + rem;
This line only knows how to create ordinary digits in the result. For (say) base 16, it would also have to be able to create hexadecimal digits A - F. One quick but obfuscated way to achieve this is to replace that line with
result[j] = "0123456789ABCDEF"[rem];
This answer is too long already, so I'm not going to get into a side discussion on how this trick works.
The second problem is hiding in the lines I added to handle negative numbers:
if(negflag) {
j--;
result[j] = '-';
}
There's no check here that there's enough room in the result array for the minus sign. If the array was just barely big enough for the converted number without the minus sign, we'll hit this part of the code with j being 0, and we'll subtract 1 from it, and fill the minus sign in to result[-1], which of course doesn't exist.
Finally, on a two's complement machine, if you pass the most negative integer, INT_MIN, in to this function, it won't work. On a 16-bit 2's complement machine, the problem number is -32768. On a 32-bit machine, it's -2147483648. The problem is that +32768 can't be represented as a signed integer on a 16-bit machine, nor will +2147483648 fit in 32 signed bits. So a rewrite of some kind will be necessary in order to achieve a perfectly general function that can also handle INT_MIN.
In order to convert a decimal number to a binary number, there is a simple recursive algorithm to apply to that number (recursive = something that is repeated until something happen):
take that number and divide by 2
take the reminder
than repeat using as current number, the original number divided by 2 (take in account that this is a integer division, so 2,5 becomes 2) until that number is different to 0
take all the reminders and read from the last to the first, and that's the binary form of that number
What that function does is exactly this
take the number and divide it by 2
takes the reminder and add it in into the variable binary multiplied by and i that each time is multiplied by 10, in order to have the first reminder as the less important digit, and the last one as the most significant digit, that is the same of take all the reminders and read them from the last to the first
save as n the n/2
and than repeat it until the current number n is different to 0
Also getch() is sometimes used in Windows in order to hold the command prompt open, but is not that recommended
getchar() stops your program in console. Maths behind function looks like this:
n=7:
7%2=1; //rem=1
7/2=3; //n=3
binary=1;
next loop
n=3:
3%2=1;
3/2=1; //n=1;
binary=11 //1 + 1* 10
final loop
n=1:
1%2=1;
1/2=0; //n=0;
binary=111 //11+1*100
This is homework, so I am not looking for a direct answer I am more-so looking for the logic behind this. I do not believe the question is stated very well for novice C devs, and I cannot find any resources to help me out here. I am new to C much more a Java guy so this may seem totally and utterly noobish. The instructions are below
$ ./mixedupecho HELLO!
.H/EmLiLxOe!dHuEpLeLcOh!oH
*
For this program, you can ignore any command-line arguments beyond the first two (including the program name itself):
$ ./mixedupecho HELLO! morestuff lalala
.H/EmLiLxOe!dHuEpLeLcOh!oH
*
Notice how "HELLO!" is shorter than "./mixedupecho", and so the program "wraps around"
and starts over again at 'H'whenever it reaches the end of the string.
*
How can you implement that? The modulo % operator is your friend here.
Spcecifically, note that "HELLO!"[5] yields '!', and "HELLO!"[6] is beyond the bounds of the array.
But "HELLO!"[6 % 6] evaluates to "HELLO!"[0], which yields 'H'.
And "HELLO!"[7 % 6] evaluates to "HELLO!"[1] ...
Below is the code I have so far. This iterates through the every character of the argv string which I get. What I don't get is how to print it off so instead of the sequence [0][0], [0][1], [0][2]... I get [0][0], [1][0], [0][1]... etc.
Can someone take a crack at explaining this to me?
int main(int argc, string argv[])
{
for(int i = 0; i < argc; i++)
{
for(int j = 0, n = strlen(argv[i]); j < n; j++)
{
printf("%c", argv[i][j]);
}
}
printf("\n");
}
THANKS SO MUCH! THIS IS DRIVING ME INSANE!
You want to keep the index i incrementing until it is equal to the index of the null terminator in the longest string. Meanwhile, you'll use the % operator to ensure that i stays within the boundaries of the shorter string.
Here's how I'd do it:
Set the initial (unsigned) lengths to -1U to avoid calculating lengths unnecessarily. I'll use LIMIT for the rest of this example as if I did #define LIMIT -1U.
Iterate through the strings, checking to ensure that argv[N][i] % len[N] is not a null terminator. If it is a null terminator and len[N] == LIMIT, set len[N] = i.
When the expression len[0] != LIMIT && len[1] != LIMIT is true, the loop ends since both strings will have the correct length, meaning all characters in each string have been enumerated.
The only thing left is printing the character for each string, which I'm sure you can handle. I would have used 0 as the initial length, except that complicates things since you can't do x % 0. The reason for unsigned length is that -1U results in an unsigned int value (e.g. 4294967295 or 65535); plain -1 results in x % -1, which makes no sense because dividing by -1 yields no remainder.
I have written a code which creates multiple random strings. But every time I print it, only the last string is printed multiple times even though different strings are created every time. Can anyone tell me what I'm doing wrong.
static const char alphanum[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" "abcdefghijklmnopqrstuvwxyz";
char s[5],*b[5] ;
int num =0;
for(int j=0;j<5;j++)
{
*b=(char*)malloc(sizeof(char*)*10);
for (int i = 0; i < 4; ++i)
{
num = rand() % (sizeof(alphanum) - 1);
s[i] = alphanum[num];
}
s[4] = 0;
printf("%s\t",s);
b[j] = s;
}
for(int j=0;j<5;j++)
printf("\n%s",b[j]);
}
Assuming that you've seeded the random number generator with, for instance, srand(time(NULL));, so that it will generate different random number sequences on each run of the program, there is one more flaw in your code:
s is a pointer to an array of characters. With the assignment b[j] = s;, you only assign b[j] the pointer (memory location) of s, but not the contents of s. Since the memory location of s does not change, all entries of b contain the same reference to the same string s, which has been changed multiple times. To copy the current content of s to b[j], use strcpy(), like this.
strcpy(b[j], s);
I think your should read the man 3 rand
In facts you have to "seed" your rand by calling void srand(unsigned int seed); one time in the beggining of your application
First of all, doing e.g. *b is the same as *(b + 0) which is the same as b[0]. That means that when you allocate memory you assign it to the same entry all the time.
Secondly, last in the loop you overwrite the pointer and make b[j] point to s, all the time. So all pointers in b will point to the same s. That's why all your strings seems to be the same.
Thirdly, you don't need to allocate dynamically in the loop, as all strings are of a fixed size. Instead declare b as an array of arrays of characters:
char b[5][5];
Then instead of assigning the pointer, you copy the string into the correct entry in b.
Lastly, and for future reference, don't cast the return of malloc.