How much memory should be malloc'ed? In other words, how would you replace XXXXXXXXXX ? The strcpy definition is:

char* strcpy(char* dest, const char* src);

/* return a freshly-malloc'd copy of s */
/* or 0 if malloc fails */
/* It is the caller's responsibility to free the returned string when done. */
char *strdup(const char *s)
{
char *s2;

s2 = malloc(XXXXXXXXXX);

if(s2 != 0) {
strcpy(s2, s);
}

return s2;
}

Question

How much memory should be malloc'ed? In other words, how would you replace XXXXXXXXXX ? The strcpy definition is:

char* strcpy(char* dest, const char* src);

/* return a freshly-malloc'd copy of s */
/* or 0 if malloc fails */
/* It is the caller's responsibility to free the returned string when done. */
char *strdup(const char *s)
{
     char *s2;
 
     s2 = malloc(XXXXXXXXXX);
 
     if(s2 != 0) {
         strcpy(s2, s);
     }
 
     return s2;
}

BlackTom AI · Accepted Answer

A C string's storage must include the terminating NUL byte, so you must allocate space for strlen(s) characters plus one extra byte for '\0'.

1. Relevant knowledge points and concepts
- strlen(s) returns the number of characters in s, not including the terminating NUL byte.
- strcpy(dest, src) copies all characters of src and the terminating NUL into dest; dest must have room for that extra byte.
- sizeof(char) is defined to be 1, so allocating (strlen(s) + 1) * sizeof(char) is equivalent to strlen(s) + 1 bytes.

2. Step-by-step reasoning
- Let n = strlen(s). That gives the count of characters excluding '\0'.
- strcpy will copy n bytes of character data plus 1 byte for the terminating '\0', for a total of n + 1 bytes needed.
- Therefore malloc must be given n + 1 bytes (or (n + 1) * sizeof(char)).

3. Calculations / intermediate steps (example)
- If s = "abc", strlen(s) = 3.
- Required bytes = strlen(s) + 1 = 3 + 1 = 4.
- malloc(4) (or malloc(4 * sizeof(char))) obtains enough space; strcpy will copy 'a','b','c','\0' into s2.

4. Why the correct answer is correct
- strlen(s)+1 explicitly allocates space for all characters plus the terminating NUL that strcpy will write. Without the +1, strcpy would write one byte past the allocated memory, causing undefined behavior.

5. Why the other options are incorrect
- "Any of these work": false—some choices are too small and lead to buffer overruns.
- strlen(s): too small by one byte; no room for '\0'.
- strlen(s)-1: even smaller; will truncate and certainly overrun for non-empty strings.

6. Conclusion
- The right expression to pass to malloc is strlen(s) + 1, which matches the provided correct answer.

类似问题

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This program will finally allocate the memory of ___ bytes that “ptr” will point to. #include<stdio.h> #include<stdlib.h> int main() { char ptr; ptr = (char)malloc(sizeof(char)4); ptr = (char)realloc(ptr,sizeof(char)*8); return 0; }

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