How can I scanf() the integer values I enter into an array until I hit enter.
I believe I can use getchar() != '\n'.
but how do I loop through the line ?
Suppose my input is 20 21 2 12 2. I want an array that has all those inputs.
What given functions could I use in order to scan them all in.
You are trying to read integers as characters so once read you need to convert it to integers.
Read the line to a buffer using fgets() then parse the input buffer to get integers.
Store the integers to the array.
The code looks like
char buf[300];
int a[5],i=0;
fgets(buf,sizeof(buf),stdin);
char *p = strtok(buf," ");
while(p != NULL)
{
char *endptr;
a[i] = strtol(p,&endptr,10);
if ((*endptr != '\0') && (isspace(*endptr) == 0))
printf("warning: invalid value detected\n");
else
i++;
p = strtok(NULL," ");
}
You can use the alternative strtol() instead of atoi() to convert string to integer.
PS: Your buf should be large enough to hold the whole line. fgets() read till newline character.
If you use getchar() you obtain digits one by one, so you need
to store them first in the buffer, and when white space comes,
you convert those digits into a number, and store it into array.
here is the explanation of the code I made for you.
1st if statement : if obtained character is a digit, store it in buf
2nd if statement : if obtained character is a white space or EOL and at least 1 digit is stored in buf, convert digits into number and store it in array a.
3rd if statement : if obtained character is not a digit or a white space or a EOL, warns users.
4th if statement : if obtained character is a EOL, end loop.
The code below works fine.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(){
#define BUFSIZE 50
#define ARRAYSIZE 5
int i,k,a[ARRAYSIZE];
char c,buf[BUFSIZE];
for(i=0,k=0;(i<BUFSIZE)&&(k<ARRAYSIZE);){
c=getchar();
if(isdigit(c)){
buf[i++] = c;
}else if((i>0) && (c==' ' || c=='\n')){
buf[i] = '\0';
a[k++] = atoi(buf);
i=0;
}else if(!(c==' ' || c=='\n')){
printf("warning : invalid value %c is detected\n",c);
i=0;
}
if(c=='\n'){
break;
}
}
printf("input :");
for(i=0;i<ARRAYSIZE;i++){
printf("%d, ",a[i]);
}
printf("\n");
}
Related
I am learning C and I am trying to use fgets() and strtol() to accept user input and just take the first character. I am going to make a menu that will allow a user to select options 1-3 and 4 to exit. I want each option to only be selected if '1', '2', '3', or '4' are selected. I don't want 'asdfasdf' to work. I also don't want '11212' to select the first option since it starts with a 1. I created this code so far while I started testing and for some reason this loops over the question and supplies a 0 to the input.
#include <stdio.h>
#include <stdlib.h>
int main() {
char a[2];
long b;
while(1) {
printf("Enter a number: ");
fgets(a, 2, stdin);
b = strtol(a, NULL, 10);
printf("b: %d\n", b);
}
return 0;
}
output
Enter a number: 3
b: 3
Enter a number: b: 0
Enter a number:
It should be:
Enter a number: 3
b: 3
Enter a number: 9
b: 9
You need to have enough room for the '\n' to be read or else it will be left in the input buffer and the next iteration it will be read immediately and thus make fgets() return with an empty string and hence strtol() returns 0.
Read fgets()'s documentation, it reads until a '\n' or untill the buffer is full. So the first time, it stops because it has no more room to store characters, and then the second time it still has to read '\n' and it stops immediately.
A possible solution is to increase the buffer size, so that the '\n' is read and stored in it.
Another solution, is to read all remaining characters after fgets().
The second solution could be cleanly implemented by reading one character at a time instead, since you are just interested in the first character you can discard anything else
#include <stdio.h>
#include <stdlib.h>
int main()
{
int chr;
while (1) {
// Read the next character in the input buffer
chr = fgetc(stdin);
// Check if the value is in range
if ((chr >= '0') && (chr <= '9')) {
int value;
// Compute the corresponding integer
value = chr - '0';
fprintf(stdout, "value: %d\n", value);
} else {
fprintf(stderr, "unexpected character: %c\n", chr);
}
// Remove remaining characters from the
// input buffer.
while (((chr = fgetc(stdin)) != '\n') && (chr != EOF))
;
}
return 0;
}
I'm making a program in which i ask for the username name, and i'd like to only accept strings with valid characters only (alphabetic).
I found that i can either use
do{
//since scanf returns the number of currectly input
if(scanf("%s", &name) == 1)
break;
else printf("Please enter a valid name.\n);
}while(1);
or
do{
check = 0;
scanf("%s", &name);
for(i=0; i<strlen(name; i++){
//since isalpha() returns != 0 if it's a letter
if(isalpha(name[i]) == 0){
printf("Invalid character. Please enter a valid name.\n");
check = 1;
break;
}
}
}while(check == 1);
But i'm not sure if any of those work, and what would be better to check if there isn't anything except alphabetic letters.
Also though about making all input letters (after this verification) on lower case and make the first letter upper case with
//all to lower except the first letter
for(i=1; i<strlen(name); i++){
name[i] = tolower(name[i]);
}
//first letter to upper
name[0] = toupper(name[i]);
x=1;
while(name[x] != '\0'){
//if the letter before is a white space, even the first letter, it should place the first letter of a name upper
if(name[x-1] == ' ')
name[x] = toupper(name[x]);
x++;
}
Would this work?
if(scanf("%s", &name) reads in all non-white-space, not just letters, into name and does not return if input is only "\n".
if(isalpha(name[i]) == 0){ loop is not bad, but scanf("%s", &name) still does not return if input is only "\n" or just white-space.
for(i=1; i<strlen(name); i++) name[i] = tolower(name[i]) works to make all following letters lower case, but if inefficient as code repeatedly calculates the string length.
Separate reading data and parsing data. Use fgets() to read the data and various code to test the data for correctness.
char buf[200];
fgets(buf, sizeof buf, stdin);
int n = 0;
// Skip leading white-space
// Look for A-Z, a-z or space (like a space between first & last)
// Skip white-space like \n
// Save into 'n' the current scan position
sscanf(buf, " %*[A-Za-z ] %n", &n);
if (n > 0 && buf[n] == '\0') Success(); // #user3121023
Should code need to rid buf of a potential trailing "\n", suggest:
buf[strcspn(buf, "\n")] = 0;
Let's look at each option.
First option:
do {
//since scanf returns the number of currectly input
if(scanf("%s", &name) == 1)
break;
else printf("Please enter a valid name.\n");
} while(1);
This won't quite work the way you expect. First off, what exactly is name? I'm almost sure that you want scanf("%s", name) instead (name instead of &name), unless you declared it as char name;, which would be catastrophic anyway.
Anyway, the problem I see with this approach is that you don't really validate the string. Read the man page section about %s:
s - Matches a sequence of non-white-space characters; the next pointer
must be a pointer to character array that is long enough to hold
the input sequence and the terminating null byte ('\0'), which is
added automatically. The input string stops at white space or at the
maximum field width, whichever occurs first.
Nothing says that the string is composed of alphabetic characters only.
Second option:
do{
check = 0;
scanf("%s", &name);
for(i=0; i<strlen(name); i++){
//since isalpha() returns != 0 if it's a letter
if(isalpha(name[i]) == 0){
printf("Invalid character. Please enter a valid name.\n");
check = 1;
break;
}
}
}while(check == 1);
Again, you probably want name rather than &name. You also shouldn't be calling strlen() in the for loop condition, because it's inefficient (strlen() is O(n)). A smart compiler may optimize it away, but it's hard for the compiler to know when it is safe to do so. Just call strlen() before the loop and store the result in a variable.
isalpha() expects an integer as an argument, which is expected to be either EOF or an unsigned char converted to int. Again, you don't show the declaration for name, but assuming that it is a character array, you should cast name[i] to unsigned char before calling isalpha(), so that you don't get any sign extension surprises:
if (isalpha((unsigned char) name[i]) == 0) { /* ... */ }
In fact, gcc nowadays will most likely give you a warning if you call any of the ctype family macros / functions with a plain char. The macros are deliberately written in such a way that a warning is shown, precisely because this is a common mistake. It is implementation-defined whether a plain char is signed or unsigned. You would get problems in a platform with signed chars because of sign extension (this is because typically, things like isalpha() are implemented using lookup tables, and extending the sign yields a negative number that would index the lookup table with a negative index - Oops!)
Other than this, this approach seems ok to me.
A third, maybe better option:
Since you mentioned fgets(), I think you could do this easily by combining fgets() with sscanf(). First, you read a line with fgets(). Then, you use sscanf() to match a string consisting of only characters in the range [a-zA-Z]. This can be done with the format specifier %[a-zA-Z]s. Then, you just have to check if this matched the entire line. Here's a working program:
#include <stdio.h>
#include <string.h>
int main(void) {
static char buf[512];
static char name[512];
int is_valid = 0;
while (!is_valid) {
fgets(buf, sizeof(buf), stdin);
size_t line_len = strlen(buf);
if (line_len > 0 && buf[line_len-1] == '\n') {
buf[line_len-1] = '\0';
line_len--;
}
int n = 0;
if (sscanf(buf, " %[a-zA-Z] %n", name, &n) == 1 && buf[n] == '\0') {
is_valid = 1;
} else {
printf("Please enter a valid name.\n");
}
}
printf("Name: %s\n", buf);
return 0;
}
Make sure your buffers are large enough; this code is vulnerable to buffer overflow for arbitrarily long names / lines.
Now let's see the code to make the first letter upper case:
//all to lower except the first letter
for(i=1; i<strlen(name); i++){
name[i] = tolower(name[i]);
}
//first letter to upper
name[0] = toupper(name[i]);
x=1;
while(name[x] != '\0'){
//if the letter before is a white space, even the first letter, it should place the first letter of a name upper
if(name[x-1] == ' ')
name[x] = toupper(name[x]);
x++;
}
Again, remove strlen() from the loop condition. toupper() and tolower() also expect an int as an argument representing either EOF or an unsigned char converted to int. You should cast it to unsigned char to avoid problems with possible sign extension, as I said earlier with the other example.
This is wrong:
//first letter to upper
name[0] = toupper(name[i]);
It should be:
//first letter to upper
name[0] = toupper(name[0]);
(The argument to toupper() is name[0], not name[i]).
Finally, this is useless:
x=1;
while(name[x] != '\0'){
//if the letter before is a white space, even the first letter, it should place the first letter of a name upper
if(name[x-1] == ' ')
name[x] = toupper(name[x]);
x++;
}
%s will never give you a string with whitespaces (refer to the manpage quote I pasted above).
Assuming that you want your name to have only characters a through z or A through Z then you could use this function
//Returns 1 if non alphabetic character is found, 0 otherwise
int NonAlphaCharsFound(char *name)
{
int FoundNonChar = 0;
int i, nameLength;
nameLength = strlen(name);
for(i = 0; i < nameLength; i++)
{
if((name[i] >= 'a' && name[i] <= 'z') || (name[i] >= 'A' && name[i] <= 'Z') || name[i] == ' ')
{
//do nothing if it's an alphabect character
//name[i] == ' ' is to allow for spaces if you want spaces in the name
}
else
{
FoundNonChar = 1;
break;
}
}
return FoundNonChar;
}
I'm having trouble printing each word in a separate line from an input string in C. The question from the assignment I'm doing states:
Take a sentence as input and print its words in separate lines.
My Code:
#include<stdio.h>
int main()
{
int i;
char s[100];
scanf("%s", s);
for(i=0; s[i]!='\0'; i++)
{
printf("%c", s[i]);
if(s[i]==' ')
{
printf("\n");
}
}
}
Any help would be appreciated.
In your code,
printf("%s", s[i]);
is wrong. Change it to
printf("%c", s[i]);
as, you're trying to print a char value. The conversion specifier for a char is %c.
Note: Always remember, using wrong conversion specifier will lead to undefined behaviour.
Also, while scan()-ing with %s, you cannot read the whole space-delimited input as a single string. From the man page,
%s
Matches a sequence of non-white-space characters; the next pointer must be a pointer to character array that is long enough to hold the input sequence and the terminating null byte ('\0'), which is added automatically. The input string stops at white space or at the maximum field width, whichever occurs first.
You need to use fgets() to do the job.
That said,
Indent your code properly, make it human-readable.
Chnage scanf("%s", s); to scanf("99%s", s); to avoid possible buffer overflow by putting longer input string than 99 chars.
the proper signature for main() is int main(void).
Rookie, using line-oriented input like fgets or getline is, in general, the proper way to read a line of text. However, when doing simple splitting on a single character, reading a character at a time can be advantageous.
In your case if your task is to read a sentence up to 100 characters and print the words of the sentence out on separate lines, then there is no reason to read the sentence into an array and store the words. You can simply read/print each character until a space is read, then print a newline instead of the space. The reading/printing continues until you reach 100 chars, encounter a newline or EOF:
#include <stdio.h>
#define MAXC 100
int main(void) {
int c = 0;
size_t n = 0;
printf ("\n Enter a sentence.\n\n input: ");
/* read up to 100 characters from stdin, print each word on a line */
while (n < MAXC && (c = getchar ()) != EOF && c != '\n')
{
if (c == ' ')
printf ("\n");
else
printf ("%c", c);
n++;
}
printf ("\n");
if (n == MAXC) /* read and discard remaining chars in stdin */
while ((c = getchar ()) != '\n' && c != EOF);
return 0;
}
Use/Output
$ ./bin/getchar_print_nl_space
Enter a sentence.
input: This is a sentence to split into words.
This
is
a
sentence
to
split
into
words.
Note: if you were going to store all characters, up to 100 (meaning 99 chars and 1 null-terminator), you would need to adjust the length check to n < MAXC - 1 and then null-terminate the array:
char s[MAXC] = {0};
/* read up to 99 characters from stdin into s */
while (n < MAXC - 1 && (c = getchar ()) != EOF && c != '\n')
s[n++] = c;
s[n] = '\0'; /* null-terminate after last character */
if (n == MAXC - 1) /* read and discard remaining chars in stdin */
while ((c = getchar ()) != '\n' && c != EOF);
You would then repeat the logic checking for a space and printing a newline in a for loop:
for (c = 0; c < n; c++)
if (s[c] == ' ')
printf ("\n");
else
printf ("%c", s[c]);
Understanding both manner of input, character-oriented input and line-oriented input will save you time allowing you to match the correct tool to the situation. Here, there is no "more correct" or "less correct" approach, just different ways of doing it.
I think one more way to do this work in a better way is as following.
#include <stdio.h>
#include <string.h>
#define MAX_CHAR 100
int main() {
char s[100],*c;
int i = 0;
scanf("%[^\n]", s);
//Write your logic to print the tokens of the sentence here.
for ( c = s; *c != (int)NULL; c++){
if ( *c == ' '){
*c = '\n';
}
}
printf("%s",s);
return 0;
}
Below code is the answer.
Program also calculates number of space/char and new line.
http://cprograming-char-operation.blogspot.com/2018/07/for-given-statement-print-word-in-each.html
/* Program 1_12 */
/* Count number of line, space and char */
/* Replace a char with specific newline */
/* Add blank space in first input */
#include<stdio.h>
int main()
{
int c,nl,nc,ns,nt;
nl=nc=ns=nt=0;
int d,r, prevd, prevr;
printf("Enter which char to replace :: ");
/* prev is stored before of \n */
while((d = getchar()) != '\n' && (prevd = d));
d = prevd;
printf("Enter word below \n");
while((c=getchar()) != EOF)
{
++nc;
if(c==' ')
++ns;
if(c=='\n')
++nl;
if(c=='\t')
++nt;
/* Replace a char with A */
if(c==d)
putchar('\n');
else
putchar(c);
}
printf("total char=%2d, newline=%2d, space=%2d tabs=%2d\n",nc,nl,ns,nt);
return 0;
}
/* Written by: Prakash Katudia <prakash.katudia#gmail.com> */
gcc ./my_code.c
./a.out
Enter which char to replace :: #space#
Enter word below
hello how are you
hello
how
are
you
#include<stdio.h>
#include<string.h>
int main()
{
char a[1000];
int i,len;
scanf("%[^\n]s",a);
len=strlen(a);
for(i=0;i<len;i++)
{
if(a[i] !=' ')
{
printf("%c", a[i]);
printf("\n");
}
}
}
In this program I have taken a dimensional character array of size[3][4],
as long as I enter a 3 characters for each row it will work well.
For example: if I enter abc abd abd I get the same output but if i enter more letters in the first or second or 3rd row I get an error.
How should I check for null character in 2 dimensional?
# include <stdio.h>
#include <conio.h>
# include <ctype.h>
void main()
{
int i=0;
char name[3][4];
printf("\n enter the names \n");
for(i=0;i<3;i++)
{
scanf( "%s",name[i]);
}
printf( "you entered these names\n");
for(i=0;i<3;i++)
{
printf( "%s\n",name[i]);
}
getch();
}
As pointed out by #SouravGhosh, you can limit your scanf with "%3s", but the problem is still there if you don't flush stdin on each iteration.
You can do this:
printf("\n enter the names \n");
for(i = 0; i < 3; i++) {
int c;
scanf("%3s", name[i]);
while ((c = fgetc(stdin)) != '\n' && c != EOF); /* Flush stdin */
}
How should I chk for null character in 2 dimensional ... [something has eaten the rest part, I guess]
You don't need to, at least not in current context.
The problem is in your approach of allocating memory and putting input into it. Your code has
char name[3][4];
if you enter more that three chars, you'll be overwriting the boundary of allocated memory [considering the space of \0]. You've to limit your scanf() using
scanf("%3s",name[i]);
Note:
change void main() to int main(). add a return 0 at the end.
always check the return value of scanf() to ensure proper input.
EDIT:
As for the logical part, you need to eat up the remainings of the input words to start scanning from the beginning of the next word.
Check the below code [Under Linux, so removed conio.h and getch()]
# include <stdio.h>
# include <ctype.h>
int main()
{
int i=0; char name[3][4];
int c = 0;
printf("\n enter the names \n");
for(i=0;i < 3;i++)
{
scanf( "%3s",name[i]);
while(1) // loop to eat up the rest of unwanted input
{ // upto a ' ' or `\n` or `EOF`, whichever is earlier
c = getchar();
if (c == ' ' || c == '\n' || c == EOF) break;
}
}
printf( "you entered these names\n");
for(i=0;i<3;i++)
{
printf( "%s\n",name[i]);
}
return 0;
}
(Cringing after reading the answers to date.)
First, state the problem clearly. You want to read a line from stdin, and extract three short whitespace separated strings. The stored strings are NUL terminated and at most three characters (excluding the NUL).
#include <stdio.h>
void main(int, char**) {
char name[3][4];
printf("\n enter the names \n");
{
// Read tbe line of input text.
char line[80];
if (0 == fgets(line, sizeof(line), stdin)) {
printf("Nothing read!\n");
return 1;
}
int n_line = strlen(line);
if ('\n' != line[n_line - 1]) {
printf("Input too long!\n");
return 2;
}
// Parse out the three values.
int v = sscanf(line, "%3s %3s %3s", name[0], name[1], name[2]);
if (3 != v) {
printf("Too few values!\n");
return 3;
}
}
// We now have the three values, with errors checked.
printf("you entered these names\n%s\n%s\n%s\n",
name[0], name[1], name[2]
);
return 0;
}
you might consider something on the order of scanf( "%3s%*s",name[i]);
which should, if I recall correctly, take the first three characters (up to a whitespace) into name, and then ignore anything else up to the next white space. This will cover your long entries and it does not care what the white space is.
This is not a perfect answer as it will probably eat the middle entry of A B C if single or double character entries are mode. strtok, will separate a line into useful bits and you can then take substrings of the bits into your name[] fields.
Perhaps figuring out the entire requirement before writing code would be the first step in the process.
Ok, i'm a student in his first experiences with programmaing so be kind ;) this is the correct code to print "n" times a string on screen...
#include <stdio.h>
#include <string.h>
#define MAX 80+1+1 /* 80+\n+\0 */
int main(void)
{
char message[MAX];
int i, n;
/* input phase */
printf("Input message: ");
i = 0;
do {
scanf("%c", &message[i]);
} while (message[i++] != '\n');
message[i] = '\0';
printf("Number of repetitions: ");
scanf("%d", &n);
/* output phase */
for (i=0; i<n; i++) {
printf("%s", message);
}
return 0;
}
why in the do-while form he needs to check if message[i++] != '\n' and not just message[i] != '\n'??
The proper way to write that input loop is, in my opinion, something along the lines of:
fgets(message, sizeof message, stdin);
in other words, don't use a character-by-character loop, just use the standard library's function that reads a string terminated by newline and be done.
The do { ... } while(...) loop in your code reads characters one at a time and stores them in message. The index of the next character is one more that the index of the previous character, that's why we should increase index variable i after the current character is stored. The algorithm is:
Read the next character and store it in message[i].
If this character is '\n', exit.
Increase i and goto 1.
The expression message[i++] increments i after it was used as an index into message, so that next time we will look at the next character in the string. So, while (message[i++] != '\n') combines steps 2 and 3.
The same in for-loop:
int i;
for (i = 0; scanf("%c", &message[i]) && message[i] != '\n'; ++i);
But as #unwind pointed, it's better not to use char-by-char input.