如何读取长度未知的输入字符串？

小开

If I may suggest a safer approach:

Declare a buffer big enough to hold the string:

char user_input[255];

Get the user input in a safe way:

fgets(user_input, 255, stdin);

A safe way to get the input, the first argument being a pointer to a buffer where the input will be stored, the second the maximum input the function should read and the third is a pointer to the standard input - i.e. where the user input comes from.

Safety in particular comes from the second argument limiting how much will be read which prevents buffer overruns. Also, fgets takes care of null-terminating the processed string.

More info on that function here.

EDIT: If you need to do any formatting (e.g. convert a string to a number), you can use atoi once you have the input.

小开

Take a character pointer to store required string.If you have some idea about possible size of string then use function

char *fgets (char *str, int size, FILE* file);`

else you can allocate memory on runtime too using malloc() function which dynamically provides requested memory.

小开

最佳答案

Enter while securing an area dynamically

E.G.

#include <stdio.h>
#include <stdlib.h>


char *inputString(FILE* fp, size_t size){
//The size is extended by the input with the value of the provisional
char *str;
int ch;
size_t len = 0;
str = realloc(NULL, sizeof(*str)*size);//size is start size
if(!str)return str;
while(EOF!=(ch=fgetc(fp)) && ch != '\n'){
str[len++]=ch;
if(len==size){
str = realloc(str, sizeof(*str)*(size+=16));
if(!str)return str;
}
}
str[len++]='\0';


return realloc(str, sizeof(*str)*len);
}


int main(void){
char *m;


printf("input string : ");
m = inputString(stdin, 10);
printf("%s\n", m);


free(m);
return 0;
}

小开

With the computers of today, you can get away with allocating very large strings (hundreds of thousands of characters) while hardly making a dent in the computer's RAM usage. So I wouldn't worry too much.

However, in the old days, when memory was at a premium, the common practice was to read strings in chunks. fgets reads up to a maximum number of chars from the input, but leaves the rest of the input buffer intact, so you can read the rest from it however you like.

in this example, I read in chunks of 200 chars, but you can use whatever chunk size you want of course.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


char* readinput()
{
#define CHUNK 200
char* input = NULL;
char tempbuf[CHUNK];
size_t inputlen = 0, templen = 0;
do {
fgets(tempbuf, CHUNK, stdin);
templen = strlen(tempbuf);
input = realloc(input, inputlen+templen+1);
strcpy(input+inputlen, tempbuf);
inputlen += templen;
} while (templen==CHUNK-1 && tempbuf[CHUNK-2]!='\n');
return input;
}


int main()
{
char* result = readinput();
printf("And the result is [%s]\n", result);
free(result);
return 0;
}

Note that this is a simplified example with no error checking; in real life you will have to make sure the input is OK by verifying the return value of fgets.

Also note that at the end if the readinput routine, no bytes are wasted; the string has the exact memory size it needs to have.

小开

I've seen only one simple way of reading an arbitrarily long string, but I've never used it. I think it goes like this:

char *m = NULL;
printf("please input a string\n");
scanf("%ms",&m);
if (m == NULL)
fprintf(stderr, "That string was too long!\n");
else
{
printf("this is the string %s\n",m);
/* ... any other use of m */
free(m);
}

The m between % and s tells scanf() to measure the string and allocate memory for it and copy the string into that, and to store the address of that allocated memory in the corresponding argument. Once you're done with it you have to free() it.

This isn't supported on every implementation of scanf(), though.

As others have pointed out, the easiest solution is to set a limit on the length of the input. If you still want to use scanf() then you can do so this way:

char m[100];
scanf("%99s",&m);

Note that the size of m[] must be at least one byte larger than the number between % and s.

If the string entered is longer than 99, then the remaining characters will wait to be read by another call or by the rest of the format string passed to scanf().

Generally scanf() is not recommended for handling user input. It's best applied to basic structured text files that were created by another application. Even then, you must be aware that the input might not be formatted as you expect, as somebody might have interfered with it to try to break your program.

小开

Safer and faster (doubling capacity) version:

char *readline(char *prompt) {
size_t size = 80;
char *str = malloc(sizeof(char) * size);
int c;
size_t len = 0;
printf("%s", prompt);
while (EOF != (c = getchar()) && c != '\r' && c != '\n') {
str[len++] = c;
if(len == size) str = realloc(str, sizeof(char) * (size *= 2));
}
str[len++]='\0';
return realloc(str, sizeof(char) * len);
}

小开

Read directly into allocated space with fgets().

Special care is need to distinguish a successful read, end-of-file, input error and out-of memory. Proper memory management needed on EOF.

This method retains a line's '\n'.

#include <stdio.h>
#include <stdlib.h>


#define FGETS_ALLOC_N 128


char* fgets_alloc(FILE *istream) {
char* buf = NULL;
size_t size = 0;
size_t used = 0;
do {
size += FGETS_ALLOC_N;
char *buf_new = realloc(buf, size);
if (buf_new == NULL) {
// Out-of-memory
free(buf);
return NULL;
}
buf = buf_new;
if (fgets(&buf[used], (int) (size - used), istream) == NULL) {
// feof or ferror
if (used == 0 || ferror(istream)) {
free(buf);
buf = NULL;
}
return buf;
}
size_t length = strlen(&buf[used]);
if (length + 1 != size - used) break;
used += length;
} while (buf[used - 1] != '\n');
return buf;
}

Sample usage

int main(void) {
FILE *istream = stdin;
char *s;
while ((s = fgets_alloc(istream)) != NULL) {
printf("'%s'", s);
free(s);
fflush(stdout);
}
if (ferror(istream)) {
puts("Input error");
} else if (feof(istream)) {
puts("End of file");
} else {
puts("Out of memory");
}
return 0;
}

小开

I know that I have arrived after 4 years and am too late but I think I have another way that someone can use. I had used getchar() Function like this:-

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


//I had putten the main Function Bellow this function.
//d for asking string,f is pointer to the string pointer
void GetStr(char *d,char **f)
{
printf("%s",d);


for(int i =0;1;i++)
{
if(i)//I.e if i!=0
*f = (char*)realloc((*f),i+1);
else
*f = (char*)malloc(i+1);
(*f)[i]=getchar();
if((*f)[i] == '\n')
{
(*f)[i]= '\0';
break;
}
}
}


int main()
{
char *s =NULL;
GetStr("Enter the String:- ",&s);
printf("Your String:- %s \nAnd It's length:- %lu\n",s,(strlen(s)));
free(s);
}

here is the sample run for this program:-

Enter the String:- I am Using Linux Mint XFCE 18.2 , eclispe CDT and GCC7.2 compiler!!
Your String:- I am Using Linux Mint XFCE 18.2 , eclispe CDT and GCC7.2 compiler!!
And It's length:- 67

小开

i also have a solution with standard inputs and outputs

#include<stdio.h>
#include<malloc.h>
int main()
{
char *str,ch;
int size=10,len=0;
str=realloc(NULL,sizeof(char)*size);
if(!str)return str;
while(EOF!=scanf("%c",&ch) && ch!="\n")
{
str[len++]=ch;
if(len==size)
{
str = realloc(str,sizeof(char)*(size+=10));
if(!str)return str;
}
}
str[len++]='\0';
printf("%s\n",str);
free(str);
}

小开

There is a new function in C standard for getting a line without specifying its size. getline function allocates string with required size automatically so there is no need to guess about string's size. The following code demonstrate usage:

#include <stdio.h>
#include <stdlib.h>




int main(void)
{
char *line = NULL;
size_t len = 0;
ssize_t read;


while ((read = getline(&line, &len, stdin)) != -1) {
printf("Retrieved line of length %zu :\n", read);
printf("%s", line);
}


if (ferror(stdin)) {
/* handle error */
}


free(line);
return 0;
}