//差不多了

不需要另外用缓冲区的

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

void Pause(void);
size_t DeleteChar(char* str, char ch);

int main(void)
{
    char str[] = "123123123";
    printf_s("Deleted char count is %u, then string is: %s", DeleteChar(str, '3'), str);

    Pause();
    return EXIT_SUCCESS;
}

void Pause(void)
{
    printf_s("\nPress any key to continue...");
    _getch();
}

size_t DeleteChar(char* str, char ch)
{
    size_t count = 0;

    char* p = str;
    while (*str != '\0')
    {
        if (*str != ch)
        {
            *p++ = *str;
        }
        else
        {
            ++count;
        }
        ++str;
    }
    *p = '\0';

    return count;
}

#include<stdio.h>
#include<string.h>
main()
{
    char a[20],n,b[20];
    int i,j,m;
    gets(a);
    scanf("%c",&n);
    for(i=0;i<strlen(a);i++)
    {
        if(a[i]!=n)
        {b[j]=a[i];j++;}     
       
    }
    puts(b);
   
} 

这样子就好啦

你可以把输出重新存入字符串，不过我最开始也干过把目标字符用空格代替的事

#include<stdio.h>
#include<string.h>
main()
{
    char a[20],ch;
    int i,k,n;
    gets(a);
    ch=getwchar();
    n=strlen(a);
    for(k=0;k<n;k++)
    {
        if(a[k]==ch)
        {
            for(i=k;i<n;i++)
                a[i]=a[i+1];
            
            n--;
            k--;
        }
    }
    puts(a);
   
}

#include<stdio.h>
int main()
{
    char a[20],n;
    int j,i;
    gets(a);
    scanf("%c",&n);
    for(i=0,j=0;a[j]!=0;i++,j++)
        if(a[j]==n)i--;
        else
            a[i]=a[j];
    a[i]=0;
    puts(a);
    
}

我改了下，楼主试试。欢迎交流。