Related
I am getting used to writing eBPF code as of now and want to avoid using pointers in my BPF text due to how difficult it is to get a correct output out of it. Using strtok() seems to be out of the question due to all of the example codes requiring pointers. I also want to expand it to CSV files in the future since this is a means of practice for me. I was able to find another user's code here but it gives me an error with the BCC terminal due to the one pointer.
char str[256];
bpf_probe_read_user(&str, sizeof(str), (void *)PT_REGS_RC(ctx));
char token[] = strtok(str, ",");
char input[] ="first second third forth";
char delimiter[] = " ";
char firstWord, *secondWord, *remainder, *context;
int inputLength = strlen(input);
char *inputCopy = (char*) calloc(inputLength + 1, sizeof(char));
strncpy(inputCopy, input, inputLength);
str = strtok_r (inputCopy, delimiter, &context);
secondWord = strtok_r (NULL, delimiter, &context);
remainder = context;
getchar();
free(inputCopy);
Pointers are powerful, and you wont be able to avoid them for very long. The time you invest in learning them is definitively worth it.
Here is an example:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/**
Extracts the word with the index "n" in the string "str".
Words are delimited by a blank space or the end of the string.
}*/
char *getWord(char *str, int n)
{
int words = 0;
int length = 0;
int beginIndex = 0;
int endIndex = 0;
char currentchar;
while ((currentchar = str[endIndex++]) != '\0')
{
if (currentchar == ' ')
{
if (n == words)
break;
if (length > 0)
words++;
length = 0;
beginIndex = endIndex;
continue;
}
length++;
}
if (n == words)
{
char *result = malloc(sizeof(char) * length + 1);
if (result == NULL)
{
printf("Error while allocating memory!\n");
exit(1);
}
memcpy(result, str + beginIndex, length);
result[length] = '\0';
return result;
}else
return NULL;
}
You can easily use the function:
int main(int argc, char *argv[])
{
char string[] = "Pointers are cool!";
char *word = getWord(string, 2);
printf("The third word is: '%s'\n", word);
free(word); //Don't forget to de-allocate the memory!
return 0;
}
In this code, I want to remove the newline. It means whenever I print this variable, it should give me a string without a newline but instead, the comma should replace it. I can directly add a comma when declaring but I want a separate result.
Expecting output
This,is,a,simple,sentence
Code
#include <stdio.h>
#include <string.h>
int main(){
char str[] = "This\nis\na\nsimple\nsentence";
printf("%s\n", str);
}
A simple for loop will do it:
for (int i = 0; i < sizeof(str); i++)
if (str[i] == '\n')
str[i] = ',';
This modifies the original string, rather than creating a new one.
You can create a one-line while loop using the strchr() function from the standard library to replace each newline character with a comma:
#include <stdio.h>
#include <string.h>
int main()
{
char str[] = "This\nis\na\nsimple\nsentence";
char* np;
while ((np = strchr(str, '\n')) != NULL) *np = ',';
printf("%s\n", str);
return 0;
}
A more efficient way of searching through the string is to use the last-found position (while there is one) as the starting-point for the search in the next loop:
int main()
{
char str[] = "This\nis\na\nsimple\nsentence";
char* np = str;
while ((np = strchr(np, '\n')) != NULL) *np = ',';
printf("%s\n", str);
return 0;
}
If I don't know how long the word is, I cannot write char m[6];,
The length of the word is maybe ten or twenty long.
How can I use scanf to get input from the keyboard?
#include <stdio.h>
int main(void)
{
char m[6];
printf("please input a string with length=5\n");
scanf("%s",&m);
printf("this is the string: %s\n", m);
return 0;
}
please input a string with length=5
input: hello
this is the string: hello
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;
}
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.
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.
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;
}
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.
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);
}
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;
}
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
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.
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);
}
I have a solution using standard libraries of C and also creating a string type (alias of char*) like in C++
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char* string;
typedef struct __strstr {
char ch;
struct __strstr *next;
}Strstr;
void get_str(char **str) {
char ch, *buffer, a;
Strstr *new = NULL;
Strstr *head = NULL, *tmp = NULL;
int c = 0, k = 0;
while ((ch = getchar()) != '\n') {
new = malloc(sizeof(Strstr));
if(new == NULL) {
printf("\nError!\n");
exit(1);
}
new->ch = ch;
new->next = NULL;
new->next = head;
head = new;
}
tmp = head;
while (tmp != NULL) {
c++;
tmp = tmp->next;
}
if(c == 0) {
*str = "";
} else {
buffer = malloc(sizeof(char) * (c + 1));
*str = malloc(sizeof(char) * (c + 1));
if(buffer == NULL || *str == NULL) {
printf("\nError!\n");
exit(1);
}
tmp = head;
while (tmp != NULL) {
buffer[k] = tmp->ch;
k++;
tmp = tmp->next;
}
buffer[k] = '\0';
for (int i = 0, j = strlen(buffer)-1; i < j; i++, j--) {
a = buffer[i];
buffer[i] = buffer[j];
buffer[j] = a;
}
strcpy(*str, buffer);
// Dealloc
free(buffer);
while (head != NULL) {
tmp = head;
head = head->next;
free(tmp);
}
}
}
int main() {
string str;
printf("Enter text: ");
get_str(&str);
printf("%s\n", str);
return 0;
}
If I don't know how long the word is, I cannot write char m[6];,
The length of the word is maybe ten or twenty long.
How can I use scanf to get input from the keyboard?
#include <stdio.h>
int main(void)
{
char m[6];
printf("please input a string with length=5\n");
scanf("%s",&m);
printf("this is the string: %s\n", m);
return 0;
}
please input a string with length=5
input: hello
this is the string: hello
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;
}
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.
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.
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;
}
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.
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);
}
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;
}
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
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.
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);
}
I have a solution using standard libraries of C and also creating a string type (alias of char*) like in C++
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char* string;
typedef struct __strstr {
char ch;
struct __strstr *next;
}Strstr;
void get_str(char **str) {
char ch, *buffer, a;
Strstr *new = NULL;
Strstr *head = NULL, *tmp = NULL;
int c = 0, k = 0;
while ((ch = getchar()) != '\n') {
new = malloc(sizeof(Strstr));
if(new == NULL) {
printf("\nError!\n");
exit(1);
}
new->ch = ch;
new->next = NULL;
new->next = head;
head = new;
}
tmp = head;
while (tmp != NULL) {
c++;
tmp = tmp->next;
}
if(c == 0) {
*str = "";
} else {
buffer = malloc(sizeof(char) * (c + 1));
*str = malloc(sizeof(char) * (c + 1));
if(buffer == NULL || *str == NULL) {
printf("\nError!\n");
exit(1);
}
tmp = head;
while (tmp != NULL) {
buffer[k] = tmp->ch;
k++;
tmp = tmp->next;
}
buffer[k] = '\0';
for (int i = 0, j = strlen(buffer)-1; i < j; i++, j--) {
a = buffer[i];
buffer[i] = buffer[j];
buffer[j] = a;
}
strcpy(*str, buffer);
// Dealloc
free(buffer);
while (head != NULL) {
tmp = head;
head = head->next;
free(tmp);
}
}
}
int main() {
string str;
printf("Enter text: ");
get_str(&str);
printf("%s\n", str);
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* getfield(char* line, int num) {
char* tok = line;
char* result;
if (line)
{
do
{
if (!--num)
{
tok = strchr(line, ',');
if (tok == NULL)
{
tok = &line[strlen(line)];
}
size_t fieldlen = tok - line;
if (fieldlen)
{
result = (char*)malloc(fieldlen+1);
result[fieldlen] = '\0';
strncpy(result, line, fieldlen);
return result;
}
else
{
break;
}
}
tok = strchr(line, ',');
line = tok + 1;
} while (tok);
}
result = (char*)malloc(2);
strcpy(result, "0");
return result;
}
int main()
{
FILE* stream = fopen("data.csv", "r");
char line[1024];
char *pstr;int num1,num2,num3;
char* value1,value2,value3;
while (fgets(line, 1024, stream))
{
char* tmp = strdup(line);
value1=getfield(tmp, 1);
value2=getfield(tmp, 2);
value3=getfield(tmp, 3);
num1 =strtol(value1,&pstr,10);
num2 =strtol(value2,&pstr,10);
num3 =strtol(value3,&pstr,10)
free(value1);
free(value2);
free(value3);
printf("Fields 1,2,3 would be 1=%d 2=%d 3=%d\n", num1,num2,num3);
// NOTE strtok clobbers tmp
free(tmp);
}
}
above is my C code to read the file....
:::: data.csv ::::
10,34,30
10,33,
23,45,23
25,,45
above is my file..
here my issue is I can call the function with "num" field. so that for reading of every line I suppose to call the function 3 times.. !! so the performance is too low for the large data files.. can someone help me that I can call the function at once and It will return an array.. than I can easily store and print (e.g. for the first line array[0]=10,array[1]=34,array[2]=30 )
You could speed it up by creating a fast split function that will destroy your line (not to mention the many lurking segmentation faults and memory leaks; this code has NO error checking or freeing of resources):
#include <stdio.h>
#include <stdlib.h>
char **split(char *line, char sep, int fields) {
char **r = (char **)malloc(fields * sizeof(char*));
int lptr = 0, fptr = 0;
r[fptr++] = line;
while (line[lptr]) {
if (line[lptr] == sep) {
line[lptr] = '\0';
r[fptr] = &(line[lptr+1]);
fptr++;
}
lptr++;
}
return r;
}
int main(int argc, char **argv) {
char line[] = "some,info,in a line";
char **fields = split(line, ',', 3);
printf("0:%s 1:%s 2:%s\n", fields[0], fields[1], fields[2]);
}
result:
0:some 1:info 2:in a line
I haven't run timing test on your code, but I'll bet a nickel that the problems is using malloc(). That is SLOW.
What Bart means is that a char[] array can contain multiple strings, back-to-back. If you scan through the array as a single string once, changing all ',' characters to '\0', your last line would look like:
{ '2', '5', 0, 0, '4', '5', 0, ? rest of buffer }
^ ^ ^ !
The ^ carets below mark the positions where you'd record pointers to three strings. As you can see, they are equivalent to separate strings of "25", "", "45" in separate arrays. The ! below marks the 0 that ended the original string. Nothing beyond that has any meaning.
All this depends on being able to modify the original string in-place, probably rendering it useless for any further processing (like printing out the offending line if an invalid field is detected). However, you are already copying the original buffer for local use, so that shouldn't be a problem. I'd get rid of the malloc for that copy buffer too, by the way.
Code might look like:
while (fgets(line, 1024, stream))
{
char tmp[sizeof line]; /* this will save a malloc()/free() pair */
char *tok, *fence, *pstr;
char ch, *cp1=line, *cp2=tmp;
while (0 != (ch = *cp1++))
*cp2++ = (ch == ',') ? 0 : ch;
fence = cp2; /* remember end of string */
*fence = 0; /* and terminate final string */
tok = tmp; /* point to first token */
num1 =strtol(tok, &pstr, 10);
if (tok < fence) tok += strlen(tok) + 1;
num2 =strtol(tok,&pstr,10);
if (tok < fence) tok += strlen(tok) + 1;
num3 =strtol(tok,&pstr,10);
printf("Fields 1,2,3 would be 1=%d 2=%d 3=%d\n", num1,num2,num3);
}
Obviously you don't need a 1K buffer to handle three values, so there will be a loop to pull out the values. The if statement after the first two strtol() calls is your replacement for getfield(), which isn't needed any more.
After this is working, look at data validation. Nothing in this (or in the original) will detect invalid numbers.