I am trying to write a program in C that spits out random characters. Following instructions I found here, I wrote this program.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(void) {
srandom((unsigned) time(NULL));
printf("Tests various aspects of random\n");
char misc;
int num, index;
printf("Enter number of chars: ");
scanf("%d", &num);
printf("\n");
for (int i = 0; i < num; i++) {
index = random() % 26;
misc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"[index];
printf("%d:%s\n", index, &misc);
}
}
However, it doesn't behave as I expect. When entering a small number of characters for it to generate, like 10, it makes the expected output.
My expected output is a set of
rand_int:char
pairs printed to the screen.
Here is an example of normal operation
Tests various aspects of random
Enter number of chars:
7:H
4:E
23:X
2:C
4:E
17:R
22:W
11:L
9:J
4:E
However, if I input a large value such as 100, it outputs very strange things like:
Tests various aspects of random
Enter number of chars:
18:Sd
3:Dd
21:Vd
10:Kd
19:Td
19:Td
14:Od
7:Hd
15:Pd
22:Wd
24:Yd
22:Wd
12:Md
[rest omitted for brevity...]
So the question is, why does it behave this way?
What might be a better approach to avoid this?
The comments made by Jabberwocky and Federico klez Culloca got it right.
I was trying to print the character as a string. This was wrong and did weird things.
I needed to use:
printf("%d:%c\n", index, misc);
instead of
printf("%d:%s\n", index, &misc);
All is very simple. The program has undefined behavior. You are using the format string %s that is used to output strings.
printf("%d:%s\n", index, &misc);
However the variable misc is not a character array that contains a string. It is just a single character. So the function printf outputs all characters beyond the variable misc until a zero-terminating character is encountered.
And it seems that the variable num is allocated next to the variable misc. So the printf call outputs also bytes of the variable num that contains the value 100. If to output this value stored in a byte as an ASCII character then you will get the character 'd'.
Here is a demonstrative program.
#include <stdio.h>
int main(void)
{
char c = 100;
putchar( c );
putchar( '\n' );
return 0;
}
Its output is
d
Instead of the format %s use the format %c in the printf call. For example
printf("%d:%c\n", index, misc);
Related
I have a simple program that reads a pair of characters from a char[] array and prints each pair to the console, all on the same line - for some reason, some spurious newlines (and whitespace) are added to the output.
I've removed usage of str libs (apart from strlen) that may add newlines at the end of strings - but I am still lost as to what's happening.
The program:
#include <stdio.h>
#include <string.h>
char input[] = "aabbaabbaabbaabbaabb";
int main() {
int i;
char c[2];
size_t input_length = strlen(input);
for (i=0; i<input_length; i+=2) {
c[0] = input[i];
c[1] = input[i+1];
printf("%s", c);
}
printf("\n");
return 0;
}
Expected output:
aabbaabbabbaabbaabb
Output:
aabbaabbabb
aa
bbaabb
Why are there newlines and whitespace in the output? (Note that the 1st line has a single a towards the end - could not deduce why)
Using Apple clang version 11.0.0 (clang-1100.0.33.16), though I would doubt if that matters.
%s works properly if your string contains null character ('\0'). If it does not (just like your case), then printf function continues to print characters until it finds '\0' somewhere in memory. Remember that string in C is a character sequence terminated with '\0'. This is the reason why your code does not behave as you expected.
On the other hand, %c prints only one character so you can use:
printf("%c%c", c[0],c[1]);
If you persist in using %s, in this case you have to use %.2s. You probably already know that . shows precision in C. Precision in string means maximum number of characters that you want to print. So usage of .2 results in printing the first two characters in your string. No need to wait for '\0'!
printf("%.2s", c);
I also give #Tom Karzes's solution. You should change and add these lines:
char c[3];
c[2] = '\0';
I'm new to C and I'm trying to write a program that prints the ASCII value for every letter in a name that the user enters. I attempted to store the letters in an array and try to print each ASCII value and letter of the name separately but, for some reason, it only prints the value of the first letter.
For example, if I write "Anna" it just prints 65 and not the values for the other letters in the name. I think it has something to do with my sizeof(name)/sizeof(char) part of the for loop, because when I print it separately, it only prints out 1.
I can't figure out how to fix it:
#include <stdio.h>
int main(){
int e;
char name[] = "";
printf("Enter a name : \n");
scanf("%c",&name);
for(int i = 0; i < (sizeof(name)/sizeof(char)); i++){
e = name[i];
printf("The ASCII value of the letter %c is : %d \n",name[i],e);
}
int n = (sizeof(name)/sizeof(char));
printf("%d", n);
}
Here's a corrected, annotated version:
#include <stdio.h>
#include <string.h>
int main() {
int e;
char name[100] = ""; // Allow for up to 100 characters
printf("Enter a name : \n");
// scanf("%c", &name); // %c reads a single character
scanf("%99s", name); // Use %s to read a string! %99s to limit input size!
// for (int i = 0; i < (sizeof(name) / sizeof(char)); i++) { // sizeof(name) / sizeof(char) is a fixed value!
size_t len = strlen(name); // Use this library function to get string length
for (size_t i = 0; i < len; i++) { // Saves calculating each time!
e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
printf("\n Name length = %zu\n", strlen(name)); // Given length!
int n = (sizeof(name) / sizeof(char)); // As noted above, this will be ...
printf("%d", n); // ... a fixed value (100, as it stands).
return 0; // ALWAYS return an integer from main!
}
But also read the comments given in your question!
This is a rather long answer, feel free to skip to the end for the code example.
First of all, by initialising a char array with unspecified length, you are making that array have length 1 (it only contains the empty string). The key issue here is that arrays in C are fixed size, so name will not grow larger.
Second, the format specifier %c causes scanf to only ever read one byte. This means that even if you had made a larger array, you would only be reading one byte to it anyway.
The parameter you're giving to scanf is erroneous, but accidentally works - you're passing a pointer to an array when it expects a pointer to char. It works because the pointer to the array points at the first element of the array. Luckily this is an easy fix, an array of a type can be passed to a function expecting a pointer to that type - it is said to "decay" to a pointer. So you could just pass name instead.
As a result of these two actions, you now have a situation where name is of length 1, and you have read exactly one byte into it. The next issue is sizeof(name)/sizeof(char) - this will always equal 1 in your program. sizeof char is defined to always equal 1, so using it as a divisor causes no effect, and we already know sizeof name is equal to 1. This means your for loop will only ever read one byte from the array. For the exact same reason n is equal to 1. This is not erroneous per se, it's just probably not what you expected.
The solution to this can be done in a couple of ways, but I'll show one. First of all, you don't want to initialize name as you do, because it always creates an array of size 1. Instead you want to manually specify a larger size for the array, for instance 100 bytes (of which the last one will be dedicated to the terminating null byte).
char name[100];
/* You might want to zero out the array too by eg. using memset. It's not
necessary in this case, but arrays are allowed to contain anything unless
and until you replace their contents.
Parameters are target, byte to fill it with, and amount of bytes to fill */
memset(name, 0, sizeof(name));
Second, you don't necessarily want to use scanf at all if you're reading just a byte string from standard input instead of a more complex formatted string. You could eg. use fgets to read an entire line from standard input, though that also includes the newline character, which we'll have to strip.
/* The parameters are target to write to, bytes to write, and file to read from.
fgets writes a null terminator automatically after the string, so we will
read at most sizeof(name) - 1 bytes.
*/
fgets(name, sizeof(name), stdin);
Now you've read the name to memory. But the size of name the array hasn't changed, so if you used the rest of the code as is you would get a lot of messages saying The ASCII value of the letter is : 0. To get the meaningful length of the string, we'll use strlen.
NOTE: strlen is generally unsafe to use on arbitrary strings that might not be properly null-terminated as it will keep reading until it finds a zero byte, but we only get a portable bounds-checked version strnlen_s in C11. In this case we also know that the string is null-terminated, because fgets deals with that.
/* size_t is a large, unsigned integer type big enough to contain the
theoretical maximum size of an object, so size functions often return
size_t.
strlen counts the amount of bytes before the first null (0) byte */
size_t n = strlen(name);
Now that we have the length of the string, we can check if the last byte is the newline character, and remove it if so.
/* Assuming every line ends with a newline, we can simply zero out the last
byte if it's '\n' */
if (name[n - 1] == '\n') {
name[n - 1] = '\0';
/* The string is now 1 byte shorter, because we removed the newline.
We don't need to calculate strlen again, we can just do it manually. */
--n;
}
The loop looks quite similar, as it was mostly fine to begin with. Mostly, we want to avoid issues that can arise from comparing a signed int and an unsigned size_t, so we'll also make i be type size_t.
for (size_t i = 0; i < n; i++) {
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
Putting it all together, we get
#include <stdio.h>
#include <string.h>
int main() {
char name[100];
memset(name, 0, sizeof(name));
printf("Enter a name : \n");
fgets(name, sizeof(name), stdin);
size_t n = strlen(name);
if (n > 0 && name[n - 1] == '\n') {
name[n - 1] = '\0';
--n;
}
for (size_t i = 0; i < n; i++){
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
/* To correctly print a size_t, use %zu */
printf("%zu\n", n);
/* In C99 main implicitly returns 0 if you don't add a return value
yourself, but it's a good habit to remember to return from functions. */
return 0;
}
Which should work pretty much as expected.
Additional notes:
This code should be valid C99, but I believe it's not valid C89. If you need to write to the older standard, there are several things you need to do differently. Fortunately, your compiler should warn you about those issues if you tell it which standard you want to use. C99 is probably the default these days, but older code still exists.
It's a bit inflexible to be reading strings into fixed-size buffers like this, so in a real situation you might want to have a way of dynamically increasing the size of the buffer as necessary. This will probably require you to use C's manual memory management functionality like malloc and realloc, which aren't particularly difficult but take greater care to avoid issues like memory leaks.
It's not guaranteed the strings you're reading are in any specific encoding, and C strings aren't really ideal for handling text that isn't encoded in a single-byte encoding. There is support for "wide character strings" but probably more often you'll be handling char strings containing UTF-8 where a single codepoint might be multiple bytes, and might not even represent an individual letter as such. In a more general-purpose program, you should keep this in mind.
If we need write a code to get ASCII values of all elements in a string, then we need to use "%d" instead of "%c". By doing this %d takes the corresponding ascii value of the following character.
If we need to only print the ascii value of each character in the string. Then this code will work:
#include <stdio.h>
char str[100];
int x;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
printf("%d\n",str[x]);
}
}
To store all corresponding ASCII value of character in a new variable, we need to declare an integer variable and assign it to character. By this way the integer variable stores ascii value of character. The code is:
#include <stdio.h>
char str[100];
int x,ascii;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
ascii=str[x];
printf("%d\n",ascii);
}
}
I hope this answer helped you.....š
so i've been writing a program that convert a decimal number to it's boolean representation but every time i compile the return value which is a string show additional characters like pā here is the program
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main (void)
{
signed char str[256];
int dec,rest;
int i = -1;
int j;
printf ("write a decimal number : ");
scanf ("%d",&dec);
do
{
rest = dec % 2;
dec/= 2;
for (j=i;j>-1;--j)
str[j+1]=str[j];
str[0]=rest+48;
++i;
} while (dec!=0);
printf ("the binary representation of this number is %s",str);
}
the output :
write a decimal number : 156
the binary representation of this number is 10011100pā
i don't know if im missing something but i will be grateful if you guys help me
In C and C++, strings are null-terminated, this means that every valid string must end with a character with code 0. This character tells every function that is dealing with this string that it is in fact over.
In your program you create a string, signed char str[256]; and it is initially filled with random data; this means that you reserved space for 256 characters and they are all garbage, but the system does not know they are invalid. Try printing this string and see what happens.
In order to actually tell the system that your string is over after say, 8 characters, the 9th character hast to be the NUL character, or simply 0. In your code you can do it in two ways:
after the loop, assign str[i] = 0, or (even simpler)
initialize the string as signed char str[256]={0};, whiche creates the storage and fills it with nulls; after writing to the string you can be sure that the character after the last one you've written will be a NUL.
At the end of your do {} while () loop, you need to set the character after the last character in your string to 0. This is the array index of the last character you want (i) plus one. This lets printf know where your string ends. (Otherwise, how could it know?)
initialize the str variable to NUL.
void main (void)
{
signed char str[256];
int dec,rest;
int i = -1;
int j;
memset( str, '\0', sizeof(str) );
printf ("write a decimal number : ");
scanf ("%d",&dec);
do
{
rest = dec % 2;
dec/= 2;
for (j=i;j>-1;--j)
str[j+1]=str[j];
str[0]=rest+48;
++i;
} while (dec!=0);
printf ("the binary representation of this number is %s",str);
}
Whenever i declare a varible as char and scan it as string "%s" my output console crashes. Here is the code
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
int main()
{
char a[20];
int i;
printf("Enter a name ");
scanf("%s",&a);
for(i=0;i<strlen(a);i++)
{
a[i] = toupper(a[i]);
i++;
printf("%s\n",toupper(a[i]));
}
return 0;
}
The second i++; inside the for loop body may cause the index to point off-by-one then, printf("%s\n",toupper(a[i])); will be out of bound access which invokes undefined behavior.
You can remove the i++ from inside the loop body.
Next, toupper(a[i]) returns an int, which is invalid for %s format specifier, which again invokes UB.
That said,
to prevent buffer overflow from excessive long input, it's best to limit the input length with scanf(),
You don't need to pass the address of an array, just the array name will be sufficient
So, overall, you should write
scanf("%19s",a);
You have two problems in the loop where you use toupper and print.
The first is that you increment the variable i twice in the loop.
The second problem is the printf:
printf("%s\n",toupper(a[i]));
Here you ask printf to print a string, but as argument you give it a single character (actually an int, toupper returns an int). Use the "%c" format specifier to print a single character.
By the way, you don't need to call toupper when printing, as the character should already be in upper-case because of the previous assignment.
Your printing loop is mis-behaving and giving you undefined behavior:
printf("%s\n",toupper(a[i]));
is passing toupper(a[i]) to printf()'s %s format specifier, which requires a char * not a plain char. That triggers undefined behavior.
There's no point in printing each character one by one, instead convert the entire string to upper case, then print it once:
for(i = 0; a[i] != '\0'; ++i)
{
a[i] = (char) toupper((unsigned char) a[i]);
}
printf("%s\n", a);
The casts around toupper() are sometimes needed since it takes and returns int, and you want to be a bit careful about when converting to/from characters.
Notice that I factored out the call to strlen(), this might be a bit "too clever" but it's how I would expect this code to be written. If we're going to loop over the characters, there's no need to compute the length separately.
The goal of this program is to create a function which reads in a single string, user typed, command (ultimately for program to be used in conjunction with a robot) which consists of an unknown command word(stored and printed as command), and an unknown number of decimal parameters(the quantity is stored and printed as num, and the parameters are to be stored as float values in the array params). In the User input, the command and parameters will be separated by spaces. I believe my issue is with the atof function when I go to extract the decimal values from the string. What am I doing wrong? Thank you for the help!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void func(char *input, char *command, int *num, float *params);
int main()
{
char input[40]={};
char command[40]={};
int num;
float params[10];
printf("Please enter your command: ");
gets(input);
func(input,command,&num,params);
printf("\n\nInput: %s",input);
printf("\nCommand: %s",command);
printf("\n# of parameters: %d",num);
printf("\nParameters: %f\n\n",params);
return 0;
}
void func(char *input, char *command, int *num, float *params)
{
int i=0, k=0, j=0, l=0;
int n=0;
while(input[i]!=32)
{
command[i]=input[i];
i++;
}
for (k=0; k<40;k++)
{
if ((input[k]==32)&&(input[k-1]!=32))
{
n++;
}
}
*num=n;
while (j<n)
{
for (l=0;l<40;l++)
{
if((input[l-1]==32)&&(input[l]!=32))
{
params[j]=atof(input[l]);
j++;
}
}
}
}
A Sample Output Screen:
Please enter your command: Move 10 -10
Input: Move 10 -10
Command: Move
# of parameters: 2
Parameters: 0.000000
The Parameters output should, ideally, read "10 -10" for the output. Thanks!
Change atof(input[l]) to atof(input + l). input[l] is single char but you want to get substring from l position. See also strtod() function.
Other people have already remarked the problem in your code, but may I suggest that you have a look at strtod() instead?
While both atof() and strtod() discard spaces at the start for you (so you don't need to do it manually), strtod() will point you to the end of the number, so that you know where to continue:
while(j < MAX_PARAMS) // avoid a buffer overflow via this check
{
params[j] = strtod(ptr, &end); // `end` is where your number ends
if(ptr == end) // if end == ptr, input wasn't a number (say, if there are none left)
break;
// input was a number, so ...
ptr = end; // continue at end for next iteration
j++; // increment number of params
}
Do note that the above solution does not differentiate between invalid arguments (say, foo instead of 3.5) and missing ones (because we've hit the last argument). You can check for that by doing this: if(!str[strspn(str, " \t\v\r\n\f")]) --- this checks if we're at the end of string (but allowing trailing whitespace). See the second side-note for what it does.
SIDE-NOTES:
You can use ' ' instead of 32 to check for space; this has two advantages:
It is clearer to the reader (it's very clear that it's a whitespace, instead of "some magic number that happens to have meaning")
It works in non-ASCII encodings (and the standard allows other encodings, though ASCII is by far the most popular; one common encoding is EBCDIC)
For future reference, this trick can help you skip whitespace: ptr += strspn(ptr, " \t\v\r\n\f");. strspn returns the number of characters at the start of the string that match the set (in this case, one of " \t\v\r\n"). Check documentation for more info.
Example for strspn: strspn("abbcbaa", "ab"); returns 3 because you have aab (which match) before c (which doesn't).
you are trying to convert a char into a float,
params[j]=atof(input[l]);
you should get the entire word(substring) of the float.
Example, "12.01" a null terminated string with 5 characters and pass it to atof, atof("12.01") and it will return a double of 12.01.
so, you should first extract the string for each float parameter and pass it to atof
Avoid comparing character to ascii value, rather you could have use ' ' (space) directly.
Instead of using for loop with a fixed size, you can use strlen() or strnlen() to find the length of the input string.