Program running properly local display triggered a breakpoint, online display error.
We know the string str, which represents the nonstandard format of the mailbox. Among them ". Will be recorded as "dot", and "
@" will be recorded as "at".
Write a program to convert str into a usable mailbox format.
Input string str.(1<)
Input string STR.(1< =strlen(str)< =1000)
#include <stdio.h>
#include <stdlib.h>
char* solution(char m[]) {
char* result;
int i, p;
int a=1;
for (i = 1; i <=998; i++)
{
if (i > 1)
{
if (m[i] == 'd' && m[i + 1] == 'o' && m[i + 2] == 't' && i < 998 && m[i + 3] != ' ')
{
m[i] = '.';
for (p = i + 1; p < 998; p++)
m[p] = m[p + 2];
m[p + 1] = ' ';
m[p + 2] = ' ';
}
if (m[i] == 'a' && m[i + 1] == 't' && a && m[i + 2] != ' ')
{
m[i] = '@';
for (p = i + 1; p < 999; p++)
m[p] = m[p + 1];
m[p + 1] = ' ';
a--;
}
}
}
result = &m[0];
return result;
}
int main() {
char str[1000];
scanf("%s", str);
solution(str);
return 0;
}
There are problems in both places. I would like to ask the reason for the mistake.
0 Answer
if(m[i] == 'd' & & m[i + 1] == 'o' & & m[i + 2] == 't' & & i < 998 & & m[i + 3] ! = ")
if(i < 997 & & m[i] == 'd' & & m[i + 1] == 'o' & & m[i + 2] == 't' & & m[i + 3] ! = ' ')
if(m[i] == 'a' & & m[i + 1] == 't' & & a & & m[i + 2] ! = ")
if(i < 998 & & m[i] == 'a' & & m[i + 1] == 't' & & a & & m[i + 2] ! = ' ')
This answer quotes ChatGPT
The reason this code might work fine locally but fail online is that it uses Pointers to return the address of a local variable. In the solution function, it points char* result to the first address of char m[] and returns result. However, when the solution function exits, the memory space pointed to by m is freed, so the memory space pointed to by result is now undefined. This can lead to errors on some platforms.
To solve this problem, you can define a character array char result[1000] in the main function and pass it as an argument to the solution function. Modify the result array in the solution function, not the m array. This way, when the solution function exits, the result array is still defined and accessible in the main function.
Here is an example of the modified code:
#include
#include
void solution(char m[], char result[]) {
int i, p;
int a = 1;
for (i = 1; i <= 998; i++) {
if (i > 1) {
if (m[i] == 'd' && m[i + 1] == 'o' && m[i + 2] == 't' && i < 998 && m[i + 3] != ' ') {
result[i - 1] = '.';
for (p = i + 1; p < 998; p++)
result[p - 1] = m[p + 1];
result[p - 1] = ' ';
result[p] = '\0';
}
if (m[i] == 'a' && m[i + 1] == 't' && a && m[i + 2] != ' ') {
result[i - 1] = '@';
for (p = i + 1; p < 999; p++)
result[p - 1] = m[p + 1];
result[p - 1] = ' ';
result[p] = '\0';
a--;
}
}
}
}
int main() {
char str[1000], result[1000];
scanf("%s", str);
solution(str, result);
printf("%s", result);
return 0;
}
这家伙很懒，什么都没留下...
