使用strcat将字符附加到C中的字符串

Question

Hi guys I'm still really confused with pointers and I'm wondering if there's anyways to do the following without having to use sprintf:

char a[100], b[100], c[2];

//Some code that puts a string into a 

for(i = 0; i<strlen(a); i++)
{
    if(a[i] == 'C')
        strcat(b, "b");
    else if(a[i] == 'O')
        strcat(b, "a");
    else if(a[i] == 'D')
        strcat(b, "1");
    else
    {
        sprintf(c, "%s", a[i]);
        strcat(b, c);
    }
}

pretty much a for loop looping through a string(an array) and filling up another string with a character(or string) depending on what the character is, if the character ain'T C, O or D it just adds it to the other string.

• I can't seem to just do strcat(b, a[i]); and I understand that it wouldn't work because it would try strcat(char *, char) instead of char*, const char*) .
  Is there anyway I can turn it into a pointer? they still confuse me so much..and I'm new to programming in general just to low level languages...

• also what would be the best way to initialize char[] s? that are gonna be filled with a string, what I use right now is :

   char ie[30] = "" 

• Also let me know if there's any easier way to do what

I want and sorry if it's unclear this is obviously a throwaway script but the same concept is used in my script.

Thank you in advance stackoverflow :X

Answer 1

(1) One bug may be in your code:

You are commenting that Some code that puts a string into a , and I think you don't assign any string to b . so by default char b[100]; have garbage value (may not present \\0 in b ). but string concatenation function expects that b must be a string. So

strcat(b, "b");   <--will Undefined Behavior 

(2) A technique to initialize empty string:

Yes you should always initialize you variable (array) with default values like:

char a[100] = {0}, b[100] = {0}, c[2] = {0};

note: remaining elements of a half initialize array would be 0 (null), Initialize a variable assume to be good practice

(3) Yes strcat(b, a[i]); is wrong:

To concatenate string from a[i] on words into b you can do like:

strcat(b, a + i);

yes you are correct strcat(b, a[i]); is not valid indeed.

note: a[i] and (a + i) are not same, a[i] is char type, where as (a + i) is string that is type of a .

Suppose you have following string array a and value of i is 2 then:

+----+----+----+---+---+----+----+----+---+
| 'u'| 's' |'e'|'r'|'5'| '6'| '7'|'8' | 0 |  
+----+----+----+---+---+----+----+----+---+
 201   202  203 204 205 206  207   208 209 210  211
  ^          ^  
  |          |
  a         (a+i) 

So in above diagram a values is 201 and type is char [100] (assuming array is 100 in size) (a + i) also points a string from 'e' at address 203 . where as a[i] = 'e'

So you can't do strcat(b, a[i]); but strcat(b, a + i); is valid syntax.

Additionally, From @BenVoigt to concat n chars from a from i th position you can do like:

strncat(b, a+i, n);

its will append n char from a+i to b .

Answer 2

既然你要采取的一个子a一个字符长：

strncat(b, a+i, 1);

Answer 3

Initialize all char array to null so that no garbage values exists in code.

You are appending to garbaged char array.

char a[100]={0}, b[100]={0}, c[2]={0};

Now strcat() function behaves properly.

Answer 4

There are many possible ways to do as you wish. There are ways that avoid using strcat() and sprintf() altogether — see below; you can avoid sprintf() while continuing to use strcat() .

The way I'd probably do it would keep a record of where the next character is to be added to the target string, b . This will be more efficient since repeatedly using strcat() involves quadratic behaviour as you build up a string one character at a time. Also, it is generally best to avoid using strlen() in the loop condition for the same reason; it is (probably) evaluated on each iteration, so that it too leads to quadratic behaviour.

char a[100], b[100];
char *bp = b;

//Some code that puts a string into a 
size_t len = strlen(a);

for (int i = 0; i < len; i++, bp++)
{
    if (a[i] == 'C')
        *bp = 'b';
    else if(a[i] == 'O')
        *bp = 'a';
    else if(a[i] == 'D')
        *bp = '1';
    else
        *bp = a[i];
}
*bp = '\0';   // Null-terminate the string

You could also do without the pointer by using the index variable i to assign to b (as long as you only add one character to the output for each input character):

char a[100], b[100];

//Some code that puts a string into a 
size_t len = strlen(a);

for (int i = 0; i < len; i++)
{
    if (a[i] == 'C')
        b[i] = 'b';
    else if(a[i] == 'O')
        b[i] = 'a';
    else if(a[i] == 'D')
        b[i] = '1';
    else
        b[i] = a[i];
}
b[i] = '\0';   // Null-terminate the string

As long as the string in a is short enough to fit, the code shown (either version) cannot overflow b . If you sometimes added several characters to b , you'd either need to indexes ( i and j ), or you could increment the pointer bp in the first version more than once per loop, and you'd need to ensure that you don't overflow the bounds of b .

Answer 5

You seem to be confused in regards to strings . A string isn't just an array. Which book are you reading?

When you first call strcat to operate on b , b isn't guaranteed to be a string. The result is undefined behaviour. This code might seem to function correctly on your system, but if it does then that is by coincidence. I have seen code like this fail in strange ways on other systems. Fix it like this:

char a[100], b[100];

//Some code that puts a string into a
a[x] = '\0'; // <--- Null terminator is required for a to contain a "string".
             //      Otherwise, you can't pass a to strlen.

for(i = 0; i<strlen(a); i++)
{
    if(a[i] == 'C')
        b[i] = 'b';
    else if(a[i] == 'O')
        b[i] = 'a';
    else if(a[i] == 'D')
        b[i] = '1';
    else
        b[i] = a[i];
}

b[i] = '\0'; // If you don't put a null character at the end, it isn't a string.

Now, what is a string?