#include <stdio.h>
#include <stdlib.h>
void print( int b)
{
    int t=0;
    for(t=0;t<b;t++)
    {
        printf(" ");
    }
}
int main()
{
    char a;
    scanf("%c",&a);
    int b;
    if(a>=97 && a<=122 )
        b=a-96;
     else if(a>=65 && a<=90)
        b=a-64;
    else
    {printf("Invalid character!\n");return 0;}
    int i,j;
    int z;
    for(i=0;i<b;i++)
    {
        print(b-i-1);
        for(j=0;j<=i;j++)
        {
            z=65+j;
            printf("%s",&z);
        }
        for(j=i-1;j>=0;j--)
        {
            z=65+j;
            printf("%s",&z);

        }

        printf("\n");
    }
    return 0;
}
可是系统一直这样回复：
cc1plus: warnings being treated as errors
main.cpp: In function ‘int main()’:
main.cpp:30: error: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘int*’
main.cpp:35: error: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘int*’

main.cpp:30: error: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘int*’
main.cpp:35: error: format ‘%s’ expects type ‘char*’, but argument 2 has type ‘int*’

他的意思是printf("%s",&z);你这里不可以这样。因为z是一个地址，int *类型的。。不能转换为%s。char*类型

 for(i=0;i<b;i++)

 {
    print(b-i-1);
    for(j=0;j<=i;j++)
    {
        z=65+j;
        printf("%c",z);  //
    }
    for(j=i-1;j>=0;j--)
    {
        z=65+j;
        printf("%c",z);  //
    }
    printf("\n");
}