[英]Bubble Sort in C array swapping
最近,當我進入這段代碼的交換部分時,我一直遇到這個問題,所以這段代碼要做的是輸入一個數組,並使用冒泡排序方法對其進行排序。 讀取此內容的文本文件有10個數字和名稱。 像這樣:
約翰一書
馬可福音2
馬太福音2
路加福音3
艾薩克4
凱恩5
瑞安7
亞伯2
亞當9
前夕10
但是,當它打印出來時,它顯示如下:
約翰一書
馬可福音2
馬太福音2
亞伯2
亞伯3
亞伯4
亞伯5
亞伯7
亞當9
前夕10
抱歉,問題是為什么它會重復Abel,我該怎么做才能解決?
void bubbleSort (Word q){
int last = 9;
int Swapped = 1;
int i = 0;
int m = 0;
char* temp = "Hello";
printf("Temp is: %s \n", temp);
int tempA;
while (Swapped == 1){
Swapped = 0;
i = 1;
while (i < last){
printf("%d \n", i);
if (q.data[i] > q.data[i+1]) {
printf("Hello: %d \n", i);
//Copy Name of Element
temp = q.name[i];
strcpy(q.name[i], q.name[i+1]);
strcpy(q.name[i+1] , temp);
//Copy Data of corresponding element
tempA = q.data[i];
q.data[i] = q.data[i+1];
q.data[i+1] = tempA;
Swapped = 1;
}
i = i + 1;
}
last = last - 1;
}
last = 9;
while (m <= last){
printf("Name: %s, Data: %d \n", q.name[m], q.data[m]);
m++;
}
}
而不是這個:
char* temp = "Hello";
您應該這樣做:
char *temp = malloc(MAX_NAME_LEN*sizeof(char));
//check if allocation of memory was successful or handle exceptions
或這個:
char temp[MAX_NAME_LEN];
接着
strcpy(temp, "Hello");
您需要將字符串存儲到指向實際內存的臨時變量中,以便在代碼的后面部分中進行字符串交換操作時使用。
而不是這樣:
//Copy Name of Element
temp = q.name[i];//HERE you are storing a pointer to the address, not the actual string. Both q.name[i] and temp point to the same memory
strcpy(q.name[i], q.name[i+1]);//Both q.name[i] and the temp pointer, with the string at q.name[i+1]. the original string in q.name[i] is lost.
strcpy(q.name[i+1] , temp);//You are copying the same content as in q.name[i+1]. You are not copying the original content of q.name[i]
做這個:
//Copy Name of Element
strcpy(temp, q.name[i]);//HERE you are storing the actual string
strcpy(q.name[i], q.name[i+1]);//q.name[i] and temp contain distinct strings
strcpy(q.name[i+1], temp);//HERE you are copying the orginal string of q.name[i]
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