[英]Pointer to Pointer to Char Array
My function passwords
takes char**
as input. 我的功能
passwords
以char**
作为输入。 I need the function to place a specific char at a location. 我需要将特定字符放置在某个位置的功能。 My program crashes and I'm not sure what I'm doing wrong but I've narrowed it down to the pointer:
我的程序崩溃了,我不确定自己在做什么错,但是我将其范围缩小到了指针:
// The function
int passwords(int num, int line, char** letters, char** ptrPassword, int length) {
char* password = *ptrPassword;
// Later in the code
password[location] = letters[line][0];
}
Here is my call from main: 这是我的主要电话:
char** password;
password = malloc(length * sizeof(char));
location = length;
//call function
printf("Password at %d is %s \n", rank, passwords(rank, 0, letters, password, length));
I'm not very experience with pointers, could someone please assist? 我对指针的经验不是很丰富,有人可以帮忙吗?
Main: 主要:
#include<stdio.h>
#include<stdlib.h>
int passwords(int num, int line, char** letters, char** ptrPassword, int length);
int lengthOfString(char* string);
void print(char** letters, int length);
int location;
int main(){
int numCase;
scanf("%d", &numCase);
int i, length;
for(i = 0; i < numCase; i++){
scanf("%d", &length);
int j;
char** letters = malloc(length*sizeof(char*));
for(j = 0; j < length; j++){
letters[j] = calloc(26, sizeof(char));
scanf("%s", letters[j]);
}
int rank;
scanf("%d", &rank);
//print(letters, j);
char* password;
password = malloc(length * sizeof(char));
location = length;
//call recursion
printf("Password at %d is %s \n", rank, passwords(rank, 0, letters, &password, length));
}
return 0;
}
The Entire Function: 整个功能:
//recursive function
int passwords(int num, int line, char** letters, char** ptrPassword, int length){
char* password = *ptrPassword;
printf("Recursion #%d \n", line);
if(line == length-1){
printf("Line is equal to length \n");
if(num > lengthOfString(letters[line])){
printf("IF \n");
if(num % lengthOfString(letters[line]) == 0){
password[location] = letters[line][lengthOfString(letters[line])];
}
else{
password[location] = letters[line][num % lengthOfString(letters[line]) - 1];
}
printf("location: %d \n", location);
location--;
printf("Password is: %s \n", password);
}
else{
printf("ELSE \n");
if(num / lengthOfString(letters[line]) == 1){
*password[location] = letters[line][0];
}
else{
printf("Alocation: %d \n", location);
password[location] = letters[line][num / lengthOfString(letters[line])];
}
printf("Blocation: %d \n", location);
location--;
printf("Password is: %s \n", password);
}
return lengthOfString(letters[line]);
}
else{
printf("Line is not equal to length \n");
int scalar = passwords(num, ++line, letters, ptrPassword, length);
if (num > scalar){
if(num % scalar == 0){
password[location] = letters[line][lengthOfString(letters[line])];
}
else{
password[location] = letters[line][num % scalar - 1];
}
location--;
}
else{
if(num / scalar == 1){
password[location] = letters[line][0];
}
else{
password[location] = letters[line][num / lengthOfString(letters[line])];
}
location--;
}
return scalar * lengthOfString(letters[line]);
}
}
Change this line 更改此行
printf("Password at %d is %s \n", rank, passwords(rank, 0, letters, &password, length));
to 至
printf("Password at %d is %d \n", rank, passwords(rank, 0, letters, &password, length));
Since your passwords
function returns an integer, not a string ( char *
). 由于您的
passwords
函数返回整数,而不是字符串( char *
)。 The same is evident from the compiler warning as well. 从编译器警告中也可以明显看出这一点。
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