I have the following code but the result is null for all components of the structure:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _TransactionType
{
char field1[20];
char field2[20];
}TransactionType;
int main(int argc, char *argv[]) {
int i;
int numreg = 0;
char temp[12];
TransactionType *dbTransaction;
dbTransaction = (TransactionType*) calloc(10,sizeof(TransactionType));
for(i=0; i<5;i++)
{
memset(temp,0,sizeof(temp));
sprintf(temp,"%d",i);
strcpy(dbTransaction->field1, temp);
dbTransaction->field1[strlen(dbTransaction->field1)] = '\0';
strcpy(dbTransaction->field2, temp);
dbTransaction->field2[strlen(dbTransaction->field2)] = '\0';
numreg++;
dbTransaction++;
}
printf("reg = %d\n", numreg);
for (i=0; i<numreg;i++)
{
printf("dbTransaction->field1 = %s\n",(dbTransaction + i)->field1);
printf("dbTransaction->field2 = %s\n",(dbTransaction + i)->field2);
}
return 0;
}
i need to recover the structure values.
Please any kind of help will be appreciate
Thanks in advance for your help
You should add error checking and casting of calloc values is discouraged, but the reason your code doesn't work is that you advance dbTransaction pointer in your loop, but never rewind it. The prints you're making are actually of elements 5-9 of the array while you fill elements 0-4.
See the corrected code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _TransactionType
{
char field1[20];
char field2[20];
}TransactionType;
int main(int argc, char *argv[]) {
int i;
int numreg = 0;
char temp[12];
TransactionType *dbTransaction;
TransactionType *dbTransactionRoot;
dbTransaction = (TransactionType*) calloc(10,sizeof(TransactionType));
dbTransactionRoot = dbTransaction;
for(i=0; i<5;i++)
{
memset(temp,0,sizeof(temp));
sprintf(temp,"%d",i);
strcpy(dbTransaction->field1, temp);
dbTransaction->field1[strlen(dbTransaction->field1)] = '\0';
strcpy(dbTransaction->field2, temp);
dbTransaction->field2[strlen(dbTransaction->field2)] = '\0';
numreg++;
dbTransaction++;
}
printf("reg = %d\n", numreg);
for (i=0; i<numreg;i++)
{
printf("dbTransaction->field1 = %s\n",(dbTransactionRoot + i)->field1);
printf("dbTransaction->field2 = %s\n",(dbTransactionRoot + i)->field2);
}
return 0;
}
Related
i wrote some code that is supposed to find the location of a given string in an array of strings.
problem is- it doesn't give the location. it gives something else.
i understand that probably the problem has to do with the differences between the pointers that are involved- a previous version that dealt with finding the position of a letter in a word worked well.
after a lot of attempts to figure out where is the bug, i ask your help.
kindly, explain me what should be done.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int what (char * token);
main()
{
int i=0;
char string[]="jsr";
char *token;
token=&string[0];
i=what(token);
printf(" location of input is %d \n", i);
return 0;
}
int what (char * token)
{
int i=1;
char *typtbl[]={"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
char * ptr;
ptr=(char *)typtbl;
while (!(strcmp(ptr,token)==0))
{
ptr=(char *)(typtbl+i);
i++;
}
return i;
}
As pointed out, you did not design function what properly. What value should it return if your search function go through all the pointers but does not find the desired string? Typically in that case return -1 would be a choice to indicate nothing found. Also in this case, using a for loop would probably be more suitable, you can just return the index immediately instead of going through all pointers.
int what(char *token)
{
char *typtbl[] = {
"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
for( size_t i = 0; i < sizeof(typtbl)/sizeof(char*); ++i )
{
char *ptr = typtbl[i];
if(strcmp(ptr, token) == 0)
{
return i; // found something
}
}
return -1; // found nothing
}
A cleaner working version.
Main issue is in the (char *)(typtbl+i) replaced by typtbl[i] in the following code. typtbl+i is equivalent to &typtbl[i], so if my memory is good, it's a pointer on the pointer of the string and not the pointer of string itself
I added a NULL at the end of the array to be able to stop if the string is not present and return -1 to clearly say it was not found.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int what(char *token);
int main()
{
int i = 0;
char string[] = "jsr";
i = what(string);
printf(" location of input is %d \n", i);
return 0;
}
int what(char *token)
{
char *typtbl[] = {
"mov",
"cmp",
"add",
"jsr",
"not",
"clr",
"lea",
NULL
};
int i = 0;
while(typtbl[i] && !(strcmp(typtbl[i], token) == 0)) {
++i;
}
if(!typtbl[i])
i = -1;
return i;
}
char *token; token=&string[0]; was useless because string == &string[0].
A few things:
Your main function is missing its return type.
The while loop in what doesn't stop when the element isn't found. Therefore you are reading out of bounds.
This should do the work w/o pointer arithmetic.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int what (char * token);
int main(){
int i=0;
char string[]="jsr";
char *token;
token=&string[0];
i=what(token);
printf(" location of input is %d \n", i);
return 0;
}
int what (char * token){
unsigned int i=0;
char *typtbl[]={"mov",
"cmp",
"add",
"sub",
"not",
"clr",
"lea",
};
unsigned int typtbl_x_size = sizeof(typtbl)/sizeof(typtbl[0]);
char * ptr;
ptr=typtbl[i];
while (!(strcmp(ptr,token)==0)){
i += 1;
if (i >= typtbl_x_size){
printf("element not in list\n");
return -1;
}
ptr=typtbl[i];
}
return i;
}
Ive been researching all day on how to merge arrays, and make functions with variable parameters. Then it got me thinking, 'can't I combine the two?'. I came up with this function. According to my understanding it should work, but I'm getting errors. Can anyone tell me what I'm doing wrong?
#include <stdio.h>
#include <stdarg.h>
char* merge(int num, ...)
{
va_list list;
char arr[9] = {0};
char *temp;
int i;
int j;
int k=0;
va_start(list,num);
for(i=0;i<num;i++)
{
temp = va_arg(list,char[]);
j = 0;
while(temp[j] != 0x00)
{
arr[k] = temp[j];
j++;
}
k++;
}
va_end(list);
return arr;
}
int main()
{
char data_1[] = "my";
char merged_array[9] = "legs";
int n=0;
//merged_array = merge(1, data_1);
while(merged_array == 0x00)
{
printf("%s\n",merged_array[n]);
n++;
}
}
Perhaps this will help get you started:
#include <stdio.h>
#include <string.h>
#include <stdarg.h>
char* merge(char *arr, int num, ...)
{
va_list list;
int i;
va_start(list,num);
for(i=0;i<num;i++)
strcat(arr, va_arg(list,char *));
va_end(list);
return arr;
}
int main()
{
char data_1[] = "my";
char merged_array[9] = "legs";
merge(merged_array, 1, data_1);
printf("%s\n", merged_array);
return(0);
}
#include <stdio.h>
#include <stdlib.h>
int countArrayChars(char *strArray[]){
int i=0;
while (strArray[i] != '\0'){
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[]) {
char *dog[] = {"dog"};
countArrayChars(dog);
For some reason, it prints "5".
Shouldn't it print 3?
I even tried to put \0 after the "g".
You declare array of string and initialize it with dog.
char *dog[] = {"dog"};
Actually it represented as
dog[0] = "Dog"; //In your case only element index with 0.
...............
...............
dog[n] = "tiger"; //If there Have n+1 element
Hence your array size is 1. Which hold constant string dog. To access it you should use dog[0].
So without less modification you can use your code as:
int countArrayChars(char *strArray[])
{
int i=0;
while (strArray[0][i] != '\0')
{
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[])
{
char *dog[] = {"dog"};
countArrayChars(dog);
}
Or if you want to declare a string use
char *dog = "dog";
or
char dog[] = "dog";
Please try this
#include <stdio.h>
#include <stdlib.h>
int countArrayChars(char *strArray){
int i=0;
while (strArray[i] != '\0'){
i++;
}
printf("%d\n", i);
return i;
}
int main(int argc, const char * argv[]) {
char *dog[] = "dog";
countArrayChars(dog);
}
#include<stdio.h>
#include<conio.h>
#include<string.h>
char* strreverse(char*);
int main()
{
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s", rev_string);
getch();
return 0;
}
char* strreverse(char *name)
{
int length=strlen(name);
char *ptr;
char *rstr;
for(ptr=name+(length-1);ptr>=name;ptr--)
{
*rstr=*ptr;
printf("%c",rstr);
rstr++;
}
*(rstr)=NULL;
return rstr;
}
the above is my code. i tried to write a program for string reverse without using arrays. But i am not getting the output retupmoc. what is wrong in my code? how to insert null char in char*?
#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <stdlib.h>
char* strreverse(const char*);
int main(){
char *rev_string;
char *name="computer";
clrscr();
rev_string=strreverse(name);
printf("%s\n", rev_string);
free(rev_string);
getch();
return 0;
}
char* strreverse(const char *name){
int length=strlen(name);
const char *ptr;
char *ret, *rstr = malloc(length + 1);
if(ret=rstr){
for(ptr=name+length;ptr != name;){
*rstr++ = *--ptr;
}
*rstr = '\0';
}
return ret;
}
You did not allocate memory to hold your reversed string. Try
char *rstr = calloc(1, length+1);
Also it should be
printf("%c", *rstr); // dereference
*(rstr)= '\0'; // instead of NULL
Here you find sweet and short solution for string reverse:
#include<stdio.h>
#include<string.h>
int strreverse(char* , char*);
int main()
{
char rev_string[10] = {0};
char name[10]="computer";
strreverse(name, rev_string);
printf("%s\n", rev_string);
return 0;
}
int strreverse(char *name, char *rStr)
{
int i = 0;
int length = strlen(name);
while(i < length)
{
rStr[i] = name[length-i-1];
i++;
}
return 0;
}
Try to run and have fun.
I have the following code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* user;
char* passwd;
int nr;
void test()
{
int i=0;
for(i=0;i<argc;i++)
printf("Hello %s \n",user);
}
int main(int argc,char*argv[])
{
int i;
nr=argc;
for (i=0; i<argc; i++)
{
user=strdup(argv[i]);
}
test();
return 0;
}
The result is the argv[argc] on all the positions. How can I fix this? I wwant to have that test() outside the loop.
**
EDIT
**
After the ANSWERS here this is my new code, which is not working. Can anyone say why?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* user;
void test(int n)
{
int i=0;
for(i=0;i<n;i++)
printf("%s \n",user[i]);
}
int main(int argc,char*argv[])
{
user = (char*) malloc(argc*sizeof(char));
int i;
for (i=0;i<argc;i++)
{
user[i]=argv[i];
}
test(argc);
return 0;
}
You are assigning to both password and user at each iteration of the for loop. The final values you see are from the last iteration. Also, there is memory leak due to overwriting the pointers from previous strdup calls. In fact, you do not need a loop:
int main(int argc,char*argv[])
{
if(argc == 3) {
user=strdup(argv[1]);
passwd=strdup(argv[2]);
} else {
// error: usage
}
test();
return 0;
}
If you want to have multiple user/password combinations:
char *user[256], *passwd[256];
void test(int n) {
int i;
for(i=0;i<n;i++)
printf("Hello %s \n",user[i]);
}
int main(int argc,char*argv[])
{
int i;
for(i = 0; i < argc && i < 256; i+=2) {
user[i]=strdup(argv[i]);
passwd[i]=strdup(argv[i+1]);
}
test(argc);
return 0;
}
Because you overwrite the pointers user and passwd in every iteration. Hence, you'll only see the last string.
If you can tell your aim of the program, a better answer can be provided. Because I am not sure whether you want to read one user and passwd Or an array of users and passwds.
After you edit, I see you want to read an array of strings:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char** user;
// or char *user[100]; /* If you want a fix length array of pointers. Now, you dont have to malloc. /*
char* passwd;
int nr;
void test(int argc)
{
int i=0;
for(i=0;i<argc;i++)
printf("Hello %s \n",user[i]);
}
int main(int argc,char*argv[])
{
int i;
nr=argc;
user = malloc(argc*sizeof(char*));
for (i=0; i<argc; i++)
{
user[i]=strdup(argv[i]);
}
test(argc);
return 0;
}
Of course; in test() you don't use i other than a loop variable and in main() you keep overwriting the previous value of user and passwd. In effect, what you do is:
user = strdup(argv[0]); /* Note: argv[0] is the program name. */
passwd = strdup(argv[0]);
user = strdup(argv[1]);
passwd = strdup(argv[1]);
user = strdup(argv[2]);
passwd = strdup(argv[2]);
user = strdup(argv[3]);
passwd = strdup(argv[3]);
printf("%s %s \n", user, passwd);
With this information, can you fix your program?
$ cat trash.c
#include <stdio.h>
#include <string.h>
void test(FILE* stream, char* usr, char* pass) {
fprintf( stream, "%s#%s\n", usr, pass);
}
int main(int argc, char** argv) {
int i = 1;
if (argc % 2) {
while(argv[i]) {
test(stdout, argv[i], argv[i + 1]);
i += 2;
}
}
return 0;
}
$ clang trash.c
$ ./a.out user1 pass1 user2 pass2
user1#pass1
user2#pass2
$
also if you call strdup() don't forget to free memory, because strdup called malloc().