This is my target:
input: string with mixed ASCII characters (uppercase, lowercase, numbers, spaces)
output: string with only uppercase characters
I have this:
#include <stdio.h>
void csere(char s[]){
int i;
for(i=0; s[i]!='\0'; i++){
if('a'<=s[i] && s[i]<='z'){
s[i]-=32;
}
printf("%c", s[i]);
}
}
void main(){
char s[1];
scanf("%s", &s);
csere(s);
}
My problem is:
The function stops at the first 'space' character in the string.
I tried to change the s[i] != '\0' in the 'for' part for i <
strlen(s) or just for s[i], but I still get the same result.
Example: qwerty --> QWERTY, but qwe rty --> QWE
(smaller problem: The program only accepts strings with length less than 12, if i change the 1 to 0 in main function.)
Thanks for help. Sorry for bad English.
scanf only scans non-whitespace characters with the %s modifier. If you want to read everything on a string you should use fgets with stdin as the third parameter:
fgets(s, sizeof s, stdin);
If you really need to use scanf for homework or something, you should use something like:
scanf("%128[^\n]", s);
Also, take note you are not allocating enough space for the string, the fact that it has not crashed is just pure coincidence... you should allocate the space on your array:
char s[128]; // change 128 for max string size
Actually, the fgets() usage I wrote earlier would only read 1 character (including the terminator string) since you only put 1 character on the array... change the array size and it should work.
You could also just use toupper() on ctype.h, but I guess this is some kind of homework or practice.
Furthermore, if you are allowed to use pointers, this would be a shorter (and probably more performant although that'd have to be tested... compilers are good these days :-) ) way to convert to uppercase (notice though it changes your original char array, and doesn't print it, although that'd be easy to modify/add, I'll leave it to you):
void strupper(char *sptr) {
while (*sptr) {
if ((*sptr >= 'a' ) && (*sptr <= 'z')) *sptr -= 32;
sptr++;
}
}
From scanf
s
Matches a sequence of bytes that are not white-space characters. The application shall ensure that the corresponding argument is a pointer to the initial byte of an array of char, signed char, or unsigned char large enough to accept the sequence and a terminating null character code, which shall be added automatically.
This means, with %s, scanf reads a string until it encounters the first white space character. Therefore, your function converts the given string only to the first space.
To the second (smaller) problem, the array s must be large enough for the entire string given. Otherwise, you overwrite the stack space and get undefined behaviour. If you expect larger strings, you must increase the size of s, e.g.
char s[100];
Related
i got a string and a scanf that reads from input until it finds a *, which is the character i picked for the end of the text. After the * all the remaining cells get filled with random characters.
I know that a string after the \0 character if not filled completly until the last cell will fill all the remaining empty ones with \0, why is this not the case and how can i make it so that after the last letter given in input all the remaining cells are the same value?
char string1 [100];
scanf("%[^*]s", string1);
for (int i = 0; i < 100; ++i) {
printf("\n %d=%d",i,string1[i]);
}
if i try to input something like hello*, here's the output:
0=104
1=101
2=108
3=108
4=111
5=0
6=0
7=0
8=92
9=0
10=68
You have an uninitialized array:
char string1 [100];
that has indeterminate values. You could initialize the array like
char string1 [100] = { 0 };
or
char string1 [100] = "";
In this call
scanf("%[^*]s", string1);
you need to remove the trailing character s, because %[] and %s are distinct format specifiers. There is no %[]s format specifier. It should look like this:
scanf("%[^*]", string1);
The array contains a string terminated by the zero character '\0'.
So to output the string you should write for example
for ( int i = 0; string1[i] != '\0'; ++i) {
printf( "%c", string1[i] ); // or putchar( string1[i] );
putchar( '\n' );
or like
for ( int i = 0; string1[i] != '\0'; ++i) {
printf("\n %d=%c",i,string1[i]);
putchar( '\n' );
or just
puts( string1 );
As for your statement
printf("\n %d=%d",i,string1[i]);
then it outputs each character (including non-initialized characters) as integers due to using the conversion specifier d instead of c. That is the function outputs internal ASCII representations of characters.
I know that a string after the \0 character if not filled completly
until the last cell will fill all the remaining empty ones with \0
No, that's not true.
It couldn't be true: there is no length to a string. No where neither the compiler nor any function can even know what is the size of the string. Only you do. So, no, string don't autofill with '\0'
Keep in minds that there aren't any string types in C. Just pointer to chars (sometimes those pointers are constant pointers to an array, but still, they are just pointers. We know where they start, but there is no way (other than deciding it and being consistent while coding) to know where they stop.
Sure, most of the time, there is an obvious answer, that make obvious for any reader of the code what is the size of the allocated memory.
For example, when you code
char string1[20];
sprintf(string1, "hello");
it is quite obvious for a reader of that code that the allocated memory is 20 bytes. So you may think that the compiler should know, when sprinting in it of sscaning to it, that it should fill the unused part of the 20 bytes with 0. But, first of all, the compiler is not there anymore when you will sscanf or sprintf. That occurs at runtime, and compiler is at compilation time. At run time, there is not trace of that 20.
Plus, it can be more complicated than that
void fillString(char *p){
sprintf(p, "hello");
}
int main(){
char string1[20];
string1[0]='O';
string1[1]='t';
fillString(&(string1[2]));
}
How in this case does sprintf is supposed to know that it must fill 18 bytes with the string then '\0'?
And that is for normal usage. I haven't started yet with convoluted but legal usages. Such as using char buffer[1000]; as an array of 50 length-20 strings (buffer, buffer+20, buffer+40, ...) or things like
union {
char str[40];
struct {
char substr1[20];
char substr2[20];
} s;
}
So, no, strings are not filled up with '\0'. That is not the case. It is not the habit in C to have implicit thing happening under the hood. And that could not be the case, even if we wanted to.
Your "star-terminated string" behaves exactly as a "null-terminated string" does. Sometimes the rest of the allocated memory is full of 0, sometimes it is not. The scanf won't touch anything else that what is strictly needed. The rest of the allocated memory remains untouched. If that memory happened to be full of '\0' before the call to scanf, then it remains so. Otherwise not. Which leads me to my last remark: you seem to believe that it is scanf that fills the memory with non-null chars. It is not. Those chars were already there before. If you had the feeling that some other methods fill the rest of memory with '\0', that was just an impression (a natural one, since most of the time, newly allocated memory are 0. Not because a rule says so. But because that is the most frequent byte to be found in random area of memory. That is why uninitialized variables bugs are so painful: they occur only from times to times, because very often uninitialized variables are 0, just by chance, but still they are)
The easiest way to create a zeroed array is to use calloc.
Try replacing
char string1 [100];
with
char *string1=calloc(1,100);
This is my code for two functions in C:
// Begin
void readTrain(Train_t *train){
printf("Name des Zugs:");
char name[STR];
getlinee(name, STR);
strcpy(train->name, name);
printf("Name des Drivers:");
char namedriver[STR];
getlinee(namedriver, STR);
strcpy(train->driver, namedriver);
}
void getlinee(char *str, long num){
char c;
int i = 0;
while(((c=getchar())!='\n') && (i<num)){
*str = c;
str++;
i++;
}
printf("i is %d\n", i);
*str = '\0';
fflush(stdin);
}
// End
So, with void getlinee(char *str, long num) function I want to get user input to first string char name[STR] and to second char namedriver[STR]. Maximal string size is STR (30 charachters) and if I have at the input more than 30 characters for first string ("Name des Zuges"), which will be stored in name[STR], after that I input second string, which will be stored in namedriver, and then printing FIRST string, I do not get the string from the user input (first 30 characters from input), but also the second string "attached" to this, I simply do not know why...otherwise it works good, if the limit of 30 characters is respected for the first string.
Here my output, when the input is larger than 30 characters for first string, problem is in the row 5 "Zugname", why I also have second string when I m printing just first one...:
Name des Zugs:aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
i is 30
Name des Drivers:xxxxxxxx
i is 8
Zugname: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaxxxxxxxx
Drivername: xxxxxxxx
I think your issue is that your train->name is not properly terminated with '\0', as a consequence when you call printf("%s", train->name) the function keeps reading memory until it finds '\0'. In your case I guess your structure looks like:
struct Train_t {
//...
char name[STR];
char driver[STR];
//...
};
In getlinee() function, you write '\0' after the last character. In particular, if the input is more than 30 characters long, you copy the first 30 characters, then add '\0' at the 31-th character (name[30]). This is a first buffer overflow.
So where is this '\0' actually written? well, at name[30], even though your not supposed to write there. Then, if you have the structure above when you do strcpy(train->name, name); you will actually copy a 31-bytes long string: 30 chars into train->name, and the '\0' will overflow into train->driver[0]. This is the second buffer overflow.
After this, you override the train->driver buffer so the '\0' disappears and your data in memory basically looks like:
train->name = "aaa...aaa" // no '\0' at the end so printf won't stop reading here
train->driver = "xxx\0" // but there
You have an off-by-one error on your array sizes -- you have arrays of STR chars, and you read up to STR characters into them, but then you store a NUL terminator, requiring (up to) STR + 1 bytes total. So whenever you have a max size input, you run off the end of your array(s) and get undefined behavior.
Pass STR - 1 as the second argument to getlinee for the easiest fix.
Key issues
Size test in wrong order and off-by-one. ((c=getchar())!='\n') && (i<num) --> (i+1<num) && ((c=getchar())!='\n'). Else no room for the null character. Bad form to consume an excess character here.
getlinee() should be declared before first use. Tip: Enable all compiler warnings to save time.
Other
Use int c; not char c; to well distinguish the typical 257 different possible results from getchar().
fflush(stdin); is undefined behavior. Better code would consume excess characters in a line with other code.
void getlinee(char *str, long num) better with size_t num. size_t is the right size type for array sizing and indexing.
int i should be the same type as num.
Better code would also test for EOF.
while((i<num) && ((c=getchar())!='\n') && (c != EOF)){
A better design would return something from getlinee() to indicate success and identify troubles like end-of-file with nothing read, input error, too long a line and parameter trouble like str == NULL, num <= 0.
I believe you have a struct similar to this:
typedef struct train_s
{
//...
char name[STR];
char driver[STR];
//...
} Train_t;
When you attempt to write a '\0' to a string that is longer than STR (30 in this case), you actually write a '\0' to name[STR], which you don't have, since the last element of name with length STR has an index of STR-1 (29 in this case), so you are trying to write a '\0' outside your array.
And, since two strings in this struct are stored one after another, you are writing a '\0' to driver[0], which you immediately overwrite, hence when printing out name, printf doesn't find a '\0' until it reaches the end of driver, so it prints both.
Fixing this should be easy.
Just change:
while(((c=getchar())!='\n') && (i<num))
to:
while(((c=getchar())!='\n') && (i<num - 1))
Or, as I would do it, add 1 to array size:
char name[STR + 1];
char driver[STR + 1];
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.....😊
#include <stdio.h>
int main(int argc, char *argv[])
{
int i, n, m;
scanf("%d %d", &n, &m);
char s[m][n];
for (i = 0; i < m; i++) {
printf("the string --\n");
scanf("%s", s[i]);
}
for (i = 0; i < m; i++) {
printf("the strings are %s \n",s[i]);
printf("\n");
}
return 0;
}
The output is:
2 2
the string --
10
the string --
11
the strings are 1011
the strings are 11
Why is the first string 1011 instead of 10?
In C, strings are represented as a sequence of char values, terminated by a null character (0 or '\0'). This means that to store a two-character string, you need space for three characters: the two characters of string content, plus the null terminator character.
Here, you've only allocated enough space for two characters in each string, but you need space for three.
So, it reads the first string into the array s[0], but the null terminator doesn't fit, and so it overflows into the second array s[1]. Now your array of arrays s looks like this: {{'1', '0'}, {'\0', ... }}.
Then, when it reads the second string into the array s[1], it overwrites the overflowed null terminator from before. And the null terminator for the second string doesn't fit into its own array, so it overflows again into the rest of the stack. The program might crash here, or corrupt other data, because you're overflowing past the end of the array.
So now your array of arrays s ends up looking like this: {{'1', '0'}, {'1', '1'}}, followed by a '\0' somewhere after the end of the array.
When printf goes to read your first string, it prints characters until it finds a null terminator. But it doesn't find one in the first string, so it keeps going, and hits the second string. It doesn't find one there either, and continues past the end of the array. In your case, luckily a null terminator was right there, but for all we know there could be something else.
To fix this, you need to allocate an extra character per string on line 9, for the null terminator:
char s[m][n+1];
There's another problem here, however. What if your input gives you the wrong length? For example, what if your input says 2 3, i.e. that the following strings will have a length of 3, but gives you the string foobar, which is 6 characters? Your code right now would overflow the buffer when it read that string, because it doesn't ensure it's the right length.
One way to avoid this would be to use gets_s instead of sscanf() for reading the strings on line 13:
gets_s(s[i], n+1);
This will read at most n characters, so avoid crashing your program or creating a security issue. However, gets_s is a C11 function, so you may not be able to use it.
You must set column size to 3 if you insert 2 characters per string, 4 for 3 characters per string and so on.
This because string in C have a termination character ('\0') in the last position.
#include <stdio.h>
int main(int argc, char *argv[])
{
int i, n, m;
scanf("%d %d", &n, &m);
char s[m][n+1];
for (i = 0; i < m; i++) {
printf("the string --\n");
scanf("%s", s[i]);
}
for (i = 0; i < m; i++) {
printf("the strings are %s \n",s[i]);
printf("\n");
}
return 0;
}
%s expects null terminated strings as an argument. When first string is read by scanf, there is not enough space for the null terminator within the allocated memory of first string. It will goes to the space next to the allocated space. Writing to unallocated space invokes undefined behavior.
While printing the strings with %s specifier, printf write the string character by character till it finds a null terminator '\0'. Here it may be the case that both the strings 10 and 11 are stored one after another in memory, so printf writes the first string till it read the null character of second string.
Input n as 3 and you will get the desire results.
In my opninion using scanf to read strings is just pure evil.
That said the array s[m][n] is just s[m*n] of course.
That said that evil thing scanf is going to load on *s[0] 10\n\0 and on *s[1*n] or *s[2] 11 and *s will be 1011\n\0
And this is a monument to bad C coding. I guess it's just an example but if I was asked this question i would say: "Come on, get me real things"
I'm extremely new to C and am doing a few problems I found in a book I bought. What is wrong with this program?
int main (void)
{
char text[50]='\0';
scanf ("%s", text);
printf("%c", text[49]);
printf("%s", text);
return 0;
}
char text[50]='\0';
is not valid. You could skip initialising text and just declare it
char text[50];
or you could initialise its first element
char text[50]={'\0'};
You're also missing an include of stdio.h and should really check that your scanf call read a string and could give it a max length for the string
if (scanf("%49s", text) == 1)
You want to get rid of:
printf("%c", text[49]);
as you have no idea what's at that memory location if the string is less than 49 chars long.
There is a difference of single quotes and double quotes in C.
double quotes means string
single quotes means character
Line 3 will not compile because the compiler wants you to assign a string to the array of characters.
You can do
char text[50]="\0";
which in effect fills all the 50 bytes with zeros.
You could also do
char text[50]="bla";
which fills the first 3 bytes with "bla" and the rest with zeros. At least my compiler does it like that.
You could also do nothing because you anyway fill it with user input just the next statement.
char text[50];
scanf ("%s", text);
But then you have a problem. Because the very next statement will give you random output if the user has entered a string with less than 49 characters. But if you initialize, well then you output the zero byte, which is also quite useless.
The main point however is to learn the different behaviour of C when dealing with an array of characters.
int main ()
{
char text[50]={'1','2','3','4'};
printf("%c", text[1]);
printf("%c",text[0]);
getch();
return 0;
}
do like this..