表达式必须有指向C中对象类型的指针。
表达式必须有指向C中对象类型的指针。
提问于 2013-11-19 06:26:30
我试图在函数和数组中使用指针，当我在main中调用report函数时，我一直得到一个错误表达式，必须有一个指向对象类型的指针。我什么都试过了。似乎什么都没起作用。有人能告诉我我做错了什么吗？
请注意:如果没有report函数，如果我在main中单独调用其他函数，它就能工作。它不仅适用于report函数。
#include <stdio.h>
#include <conio.h>
void print(int *list, int row_count, int column_count);
void rowaverage(int *list, int row_count, int column_count);
void allaverage(int *list, int row_count, int column_count);
void largest(int *list, int row_count, int column_count);
void report(int *list, int row_count, int column_count);
int main()
{
int i = 1, row, column;
int list[3][5];
printf("Enter 3 sets of 5 integers::\n");
for (row = 0; row < 3; row++)
{
printf("Elements in the %d set are ::\n", row);
for (column = 0; column < 5; column++)
{
printf("Element No. %d is ", i++);
scanf("%d", &list[row][column]);
}
printf("\n");
i = 1;
}
printf("The elements in array are:\n");
report(&list[0][0], row, column);
getch();
return 0;
}
void print(int *list, int row_count, int column_count)
{
int column, row;
for (row = 0; row < row_count; row++)
{
for (column = 0; column < column_count; column++)
{
printf("%8d", *(list + row * column_count + column));
}
printf("\n");
}
}
void rowaverage(int *list, int row_count, int column_count)
{
int column, row;
for (row = 0; row < row_count; row++)
{
float sum = 0, count = 0;
for (column = 0; column < column_count; column++)
{
sum += *(list + row * column_count + column);
count++;
}
printf("Average of row %d is %.2f\n", row, (sum / count));
}
}
void allaverage(int *list, int row_count, int column_count)
{
int column, row;
float sum = 0, count = 0;
for (row = 0; row < row_count; row++)
{
for (column = 0; column < column_count; column++)
{
sum += *(list + row * column_count + column);
count++;
}
}
printf("Average of all elements in array is %.2f\n", (sum / count));
}
void largest(int *list, int row_count, int column_count)
{
int column = 0, row = 0;
int largest = *(list + row * column_count + column);
for (row = 0; row < row_count; row++)
{
for (column = 0; column < column_count; column++)
{
if (largest < *(list + row * column_count + column))
{
largest = *(list + row * column_count + column);
}
}
}
printf("The largest number in the array is %d\n", largest);
}
void report(int *list, int row_count, int column_count)
{
int row = 0, column = 0;
print(list[0][0], row, column);
printf("\n");
rowaverage(list[0][0], row, column);
printf("\n");
allaverage(list[0][0], row, column);
printf("\n");
largest(list[0][0], row, column);
}回答 4
Stack Overflow用户
回答已采纳
发布于 2013-11-19 06:48:19
在report函数中，删除该行并参见下面的report函数：
int row = 0, column = 0;在函数中，使用列表作为
int list[][5]呼叫列表
list不像
list[0][0]以下是完整的代码：
#include <stdio.h>
#include <stdlib.h>
void print(int list[][5], int row_count, int column_count)
{
int column, row;
for (row = 0; row < row_count; row++) {
for (column = 0; column < column_count; column++)
printf("%8d", list[row][column]);
printf("\n");
}
}
void rowaverage(int list[][5], int row_count, int column_count)
{
int column, row;
for (row = 0; row < row_count; row++) {
float sum = 0, count = 0;
for (column = 0; column < column_count; column++) {
sum += list[row][column];
count++;
}
printf("Average of row %d is %.2f\n", row, (sum / count));
}
}
void allaverage(int list[][5], int row_count, int column_count)
{
int column, row;
float sum = 0, count = 0;
for (row = 0; row < row_count; row++) {
for (column = 0; column < column_count; column++) {
sum += list[row][column];
count++;
}
}
printf("Average of all elements in array is %.2f\n", (sum / count));
}
void largest(int list[][5], int row_count, int column_count)
{
int column = 0, row = 0;
int largest = list[0][0];
for (row = 0; row < row_count; row++) {
for (column = 0; column < column_count; column++) {
if (largest < list[row][column]) {
largest = list[row][column];
}
}
}
printf("The largest number in the array is %d\n", largest);
}
void report(int list[][5], int row_count, int column_count)
{
print(list, row_count, column_count);
printf("\n");
rowaverage(list, row_count, column_count);
printf("\n");
allaverage(list, row_count, column_count);
printf("\n");
largest(list, row_count, column_count);
}
int main()
{
int i = 1, row, column;
int list[3][5];
printf("Enter 3 sets of 5 integers::\n");
for (row = 0; row < 3; row++) {
printf("Elements in the %d set are ::\n", row);
for (column = 0; column < 5; column++) {
printf("Element No. %d is ", i++);
scanf("%d", &list[row][column]);
}
printf("\n");
i = 1;
}
printf("The elements in array are:\n");
report(list, row, column);
return 0;
}Stack Overflow用户
发布于 2013-11-19 06:35:22
因此，在report函数中，您是一个指向int的指针：
int list[3][5];
report(&list[0][0], row, column);
...
void report(int *list, int row_count, int column_count)
...但是，您要使用report函数中的list作为指针指向int：
list[0][0]但它不是指向int的指针。它只是有一种int*类型。因此，"list“在报表函数中被取消两次引用。这是错误的。
如果在报告()中使用"list“作为函数调用中的参数，而不使用”：
rowaverage(list, row, column);Stack Overflow用户
发布于 2013-11-19 06:40:45
也许您可以将参数list更改为list。因为list是int的指针，而list只是int的指针。它们的类型不同，函数中的第一个参数需要一个指针。:)
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/20064650
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