Why does strcat fail to concatenate two strings? - c

Question

I need to convert a number into a string. I'm a dealer and distribute cards to players. The deck contains 40 cards. Four seeds.
#include <stdio.h>
#include <conio.h>
#include <string.h>
char *convertNum(int a); //initialization function
void main(){
char *car=convertNum(30);
printf(" number to convert 30 : %s", car);
getch();
}
char *convertNum(int a){
char *seed, *value;
switch ((a-1)/10){
case 0: seed="C"; break;
case 1: seed="Q"; break;
case 2: seed="F"; break;
case 3: seed="P"; break;
default: seed="X"; break;
}
switch(a%10){
case 0: value="10"; break;
case 1: value="01"; break;
case 2: value="02"; break;
case 3: value="03"; break;
case 4: value="04"; break;
case 5: value="05"; break;
case 6: value="06"; break;
case 7: value="07"; break;
case 8: value="08"; break;
case 9: value="09"; break;
default: value="XX"; break;
}
strcat(value, seed); /* ERROR: I need to concatenate the strings value and seed, but this isn't the right way */
return value;
}
What can I do to concatenate the strings? value and seed are pointers.

Answer 1

1.The pointers *value and *seed are uninitialized . They have no meomories. You need to initialize those pointers using malloc() or calloc() with some block of memory then it can hold values.
Hope this will work for you

Answer 2

In your function,you are assigning the pointers seed and value to string literals in the switch statements,however,what you need to do is allocate memory for these two pointers,copy strings to them using strcpy() or other standard library function,and concatenate at the end of the function.Make sure the allocated memory is big enough to hold both strings along with the '\0' character.Finally don't forget to free allocated memory when done.
#include <stdio.h>
#include <conio.h>
#include <string.h>
char *convertNum(int a);
void main(){
char *car=convertNum(30);
printf("number to convert 30 : %s\n", car);
free(car);
getch();
}
char *convertNum(int a){
char *seed = (char*)malloc(20*sizeof(char)), *value = (char*)malloc(20*sizeof(char));
switch ((a-1)/10){
case 0: strcpy(seed,"C"); break;
case 1: strcpy(seed,"Q"); break;
case 2: strcpy(seed,"F"); break;
case 3: strcpy(seed,"P"); break;
default: strcpy(seed,"X"); break;
}
switch(a%10){
case 0: strcpy(value,"10"); break;
case 1: strcpy(value,"01"); break;
case 2: strcpy(value,"02"); break;
case 3: strcpy(value,"03"); break;
case 4: strcpy(value,"04"); break;
case 5: strcpy(value,"05"); break;
case 6: strcpy(value,"06"); break;
case 7: strcpy(value,"07"); break;
case 8: strcpy(value,"08"); break;
case 9: strcpy(value,"09"); break;
default:strcpy(value,"XX"); break;
}
strcat(value, seed);
free(seed);
return value;
}

Answer 3

You are pointing value and seed to string literals in your switch statements. You are not allowed to modify string literals with strcat(), because they are stored in read-only memory.
You could use malloc() in your convertNum() function to solve your problem. But then please don't forget to call free() when you no longer need a string. However, you can also provide a statically allocated array of characters from your main() function to the convertNum() function. This makes sense here, because your result strings will always have a fixed length of 3 characters:
void main(){
// Allocate an array of four characters and initialize it with zeros.
// You need 4 characters, because strings need to be zero terminated.
char car[4] = {0};
// Pass (a pointer to) the array to your covertNum() function.
convertNum(30, car);
printf(" number to convert 30 : %s", car);
getch();
}
Then you have to modify your convertNum() function as follows:
void convertNum(int a, char* result){
char *seed, *value;
switch ((a-1)/10){
[...]
}
switch(a%10){
[...]
}
strcpy(result, value); // Copy 'value' to the beginning of the 'result' array.
strcat(result, seed); // Append 'seed' to 'value'; 'result' becomes zero-terminated.
}
This gives the following output:
number to convert 30 : 10F
In case you want your output to be F10, then you have to switch value and seed in the last two lines of the convertNum() function.